Standard antiderivatives and integration methods

Since each antiderivative of a given function is equal up to a constant, we will use the unified notation \({\displaystyle \int^x} f(t) \ dt \),meaning this family og primitives (or general antiderivative).

We will sometimes use the capital letter of a function to signify its antiderivative.

Recall on standard antiderivatives

A certain amount of antiderivative can be directly calculated by taking the reverse path of the derivative or by an integration by parts.

For others, each case must be considered as a specific case.

General integration methods

Integration by parts

$$ \forall (a,b) \in D_f^2,$$

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b - \int_{a}^b (fg') \hspace{0.2em}dt $$

Integration by substitution

Let \(\phi : t \longmapsto \phi(t) \) be a function of class \(\mathcal{C}^1\) on an interval \( J \subset f(I) \).

The integration by substitution is the transposition of the derivative of a composite function, but applied to integral calculus.

$$ \forall (a,b) \in D_f^2, \ $$

Rational fractions integration methods \( : {\displaystyle \int^x} \frac{P(t)}{Q(t)}dt \)

With a second degree denominateur only\(: {\displaystyle \int^x} \frac{1}{Q_2(t)} \ dt \)

$$ S_1(x) = \frac{1}{x^2 + px + q}$$

• \( \alpha) \) Discriminant is positive\(: \Delta > 0 \)

$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{x_1, x_2 \bigr\} \Bigr],$$

$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{(x_1 - x_2)} \ ln \left| \frac{x-x_1}{x-x_2} \right|, \hspace{2em} with \ \left \{ \begin{align*} x_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ x_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} \end{align*} \right \} \qquad (if \ \Delta = p^2 - 4q > 0) $$




In the specific case where \((p=0, \ q-1)\), we do have as a bonus an explicite definition of the \(arctanh\) function:

$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$

$$arctanh(x) = \frac{1}{2} ln \left|\frac{x+1}{x-1} \right|$$



• \( \beta) \) Discriminant is zero\(: \Delta = 0 \)

$$\forall x \in \hspace{0.05em} \biggl[ \mathbb{R} \hspace{0.05em} \backslash \Bigl \{ -\frac{p}{2} \Bigr\} \biggr],$$

$$ \int^x \frac{1}{t^2 + pt + q} \ dt = - \frac{1}{ x + \frac{p}{2} } \qquad (if \ \Delta = p^2 - 4q= 0) $$



• \( \gamma) \) Discriminant is negative\(: \Delta < 0 \)

$$\forall x \in \mathbb{R},$$

$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{ \sqrt{ q - \frac{p^2}{4} } } \ arctan \left (\frac{x + \frac{p}{2}}{ \sqrt{ q - \frac{p^2}{4} } } \right ) \qquad (if \ \Delta = p^2 - 4q< 0) $$

With a numerator of the first degree and a denominator of the second degree\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)

$$ S_2(x) = \frac{Ax + B}{x^2 + px + q}$$

$$ according \ to \ the \ result \ of \ (\Delta = p^2 - 4q) \enspace \left \{ \begin{align*} if \ \Delta > 0 \Longrightarrow \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{x_1, x_2 \bigr\} \Bigr], \\ if \ \Delta = 0 \Longrightarrow \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \left\{ - \frac{p}{2} \right\} \biggr], \\ if \ \Delta < 0 \Longrightarrow \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{x_1, x_2 \bigr\} \Bigr], \end{align*} \right \} $$

Integration methods for square roots\(: {\displaystyle \int^x } \frac{1}{\sqrt{Q(t)}} \ dt, \ {\displaystyle \int^x } \sqrt{P(t)} \ dt \)

Let \(a \in \hspace{0.05em} \mathbb{R}\) be a real number.

Integrals containing \(\sqrt{a^2 - t^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{a^2 - t^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \hspace{0.05em} \Bigl [-|a|, \hspace{0.2em} |a| \Bigr], $$

$$ \int^x \frac{dt}{\sqrt{a^2 - t^2}} = arcsin\left(\frac{x}{|a|}\right) $$




2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{a^2 - t^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigl[ \hspace{0.05em} \cup \hspace{0.05em} \Bigr]0, \hspace{0.2em} |a| \Bigr[ , $$

$$ \int^x \frac{dt}{t\sqrt{a^2 - t^2}} = \frac{1}{2a} ln\left|\sqrt{a^2 - x^2} - a \right| + \frac{1}{2a} ln\left|\sqrt{a^2 - x^2} + a\right| $$




3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{a^2 - t^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigl[ \hspace{0.05em} \cup \hspace{0.05em} \Bigr]0, \hspace{0.2em} |a| \Bigr[ ,$$

$$ \int^x \frac{dt}{t^2\sqrt{a^2 - t^2}} = - \frac{1}{a^2} \frac{\sqrt{a^2 - x^2}}{x} $$




4. Simple root\(: \sqrt{a^2 - t^2} \)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \hspace{0.05em} \Bigl [-|a|, \hspace{0.2em} |a| \Bigr], $$

$$ \int^x \sqrt{a^2 - t^2} \ dt = \frac{a^2}{2} arcsin\left(\frac{x}{|a|}\right) + \frac{x}{2} \sqrt{a^2 - x^2} $$

Integrals containing \(\sqrt{a^2 + t^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{a^2 + t^2}}\)

1. Setting down \( t = |a| \ sinh(u)\)

$$\forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \hspace{0.05em} \mathbb{R}, $$

$$ \int^x \frac{dt}{\sqrt{a^2 + t^2}} = arcsinh\left(\frac{x}{|a|} \right) $$



2. Setting down \( t = |a| \ tan(u)\)

$$\Big[ \forall (a,x) \in \hspace{0.05em} \big[\mathbb{R}^* \bigr]^2 \Bigr] \lor \Big[ (a = 0) \land (x > 0) \Bigr], $$

$$ \int^x \frac{dt}{\sqrt{a^2 + t^2}} = ln \left|\sqrt{ a^2 + x^2 } + x\right|$$




Both expressions having a common member, they are both equal up to a constant and we do obtain as a bonus an explicit definition of the \(arcsinh\) function with \(a = 1\):

$$ \forall x \in \mathbb{R}, $$

$$ arcsinh(x) = ln \left| x + \sqrt{ 1 + x^2 } \right| $$




2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{a^2 + t^2}}\)

$$\forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \hspace{0.05em} \mathbb{R}^*,$$

$$ \int^x \frac{dt}{t\sqrt{a^2 + t^2}} = \frac{1}{2a} ln\left|\sqrt{a^2 + x^2}-a\right| - \frac{1}{2a} ln \left|\sqrt{a^2 + x^2} + a\right| $$




3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{a^2 + t^2}}\)

$$\forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \hspace{0.05em} \mathbb{R}^*,$$

$$ \int^x \frac{dt}{t^2\sqrt{a^2 + t^2}} =- \frac{1}{a^2} \frac{\sqrt{a^2+x^2}}{x}$$



4. Simple root\(: \sqrt{a^2 + t^2} \)

$$\Big[ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \mathbb{R}\Bigr] \lor \Big[ (a = 0) \land (x > 0) \Bigr], $$

$$ \int^x \sqrt{a^2 + t^2} \ dt = \frac{x \ \sqrt{a^2 +x^2} + a^2 \ ln\left|\sqrt{a^2 +x^2} + x \right|}{2} $$

Integrals containing \(\sqrt{t^2 - a^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{t^2 - a^2}}\)

1. Setting down \(t = |a| \ cosh(u) \)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \hspace{0.05em} ]a, +\infty[, $$

$$ \int^x \frac{dt}{\sqrt{t^2 - a^2}} = arccosh\left( \frac{x}{a}\right) $$

2. Setting down \(t = |a| \ sec(u) \)

$$\Bigl[ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[ \Bigr] \lor \Bigl[ (x = 0) \land (x > 0) \Bigr],$$

$$\int^x \frac{dt}{\sqrt{t^2 - a^2}} = ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

Both expressions having a common member, they are both equal up to a constant and we do obtain as a bonus an explicit definition of the \(arccosh\) function with \(a = 1\):

$$ \forall x \in [1, +\infty[,$$

$$ arccosh(x) = ln \left| x + \sqrt{ x^2 - 1} \right| $$




2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{t^2 - a^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[,$$

$$ \int^x \frac{dt}{t\sqrt{t^2 - a^2}}dt = \frac{1}{a} arctan \left(\frac{\sqrt{x^2 - a^2}}{a} \right) $$

3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{t^2 - a^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, $$

$$ \int^x \frac{dt}{t^2\sqrt{t^2 - a^2}} =\frac{1}{a^2} \frac{\sqrt{ x^2 - a^2 } }{x}$$



4. Simple root\(: \sqrt{ t^2 - a^2} \)

$$ \Bigl[ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[ \Bigr] \lor \Bigl[ (x = 0) \land (x > 0) \Bigr], $$

$$ \int^x \sqrt{x^2 - a^2} \ dt = \frac{ x\sqrt{x^2 - a^2} - a^2 \ ln \left| x+ \sqrt{x^2 - a^2} \ \right| }{2} $$

Integration methods and antiderivatives recap table

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Demonstrations

Recall on standard antiderivatives

This table resumes antiderivatives directly determined from the inverse operation of the derivative, namey:

$$ f(x) = \int^x F'(t)dt \ \Longleftrightarrow \ F(x) = \int^x f(t)dt $$

General integration methods

Integration by parts

With the derivative of a product, we do have:

$$ \left ( f g\right)' = f'g + g'f $$

$$ f'g = ( f g)' - g'f $$

Now, thanks to the property of linearity of the integral,

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \int_{a}^b (fg) -\int_{a}^b (fg') \hspace{0.2em}dt $$

And,

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b -\int_{a}^b (fg') \hspace{0.2em}dt $$

Then finally,

$$ \forall (a,b) \in I^2, \enspace a < b, $$

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b - \int_{a}^b (fg') \hspace{0.2em}dt $$

Integration by substitution

Let \(f : x \longmapsto f(x)\) be a function of class \(\mathcal{C}^1\) on an interval \(I = [a,b]\).

As well, let \(\phi : t \longmapsto \phi(t) \) be a function of class \(\mathcal{C}^1\) on an interval \( J \subset f(I)\).

Condering the following integral \(I(x)\):

$$ I(x) = \int_{a}^b f(x) \hspace{0.2em}dx = F(b) - F(a) \qquad(I(x)) $$

So,

$$ I(t) = \int_{\psi(a)}^{\psi(b)} \Bigl(f \circ \phi(t)\Bigr) \ \phi'(t) \ dt \qquad(I(t)) $$

$$ I(t) = \Bigl[ f \circ \phi(t) \Bigr]_{\psi(a)}^{\psi(b)} $$

Both functions \(\phi\) and \(\psi\) being two reciprocal functions, they annihilate with each other.

$$ I(t) = F(b) - F(a) $$

Thus, both expressions \( (I(x)) \) and \( (I(t)) \) are equals.

$$ \int_{\psi(a)}^{\psi(b)} f[ \phi(t)\Bigr] \ \phi'(t) \ dt = \int_{a}^b f(x) \hspace{0.2em}dx \qquad (1) $$

In the end, transforming all bounds by performing the function \( \phi \), then \( (1)\) becomes \( (1')\):

$$ \int_{\phi(\psi(a))}^{\phi(\psi(b))} f[ \phi(t)\Bigr] \ \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(x) \hspace{0.2em}dx \qquad (1') $$

And,

$$ \int_{a}^{b} f[ \phi(t)\Bigr] \ \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(x) \hspace{0.2em}dx $$

For the sake of simplicity, we will use \(u \) as a variable, and finally,

$$ \forall (a,b) \in D_f^2,$$

Be careful not to be confused with a direct variable change:

$$ \Biggl \{ \begin{align*} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{align*} $$

in opposition to an indirect variable change:

$$ \Biggl \{ \begin{align*} t = \psi(u) \\ dt = \psi '(u) \hspace{0.2em} du \end{align*} $$

Rational fractions integration methods \( : {\displaystyle \int^x} \frac{P(t)}{Q(t)}dt \)

Generally speaking, we will always try to reduce ourselves to a whole fraction with a remainder, rather than staying with a numerator of a higher degree than the denominator.

For example,

$$ R(x) = \frac{3x^2 + 2x + 1}{x+1}$$

After the Euclidian division of \((3x^2 + 2x + 1)\) by \((x+1)\) it remains:

$$ R(x) = \ \underbrace{ 3x^2 - 3x + 5 } _\text{partie entière} \ - \ \underbrace{ \frac{5}{x+1}} _\text{reste}$$

Which now allows us to easily integrate it:

$$ \int^x R(t) \ dt = \int^x \Bigl( 3t^2 - 3t + 5 \Bigr)\ dt - \int^x \frac{5}{t+1}\ dt $$

$$ \int^x R(t) \ dt = x^3 - \frac{3t^2}{2} + 5x -5ln\bigl|t+1\bigr|$$

Similarly, when we have a polynomial of type \(ax^2 + bx + c\), we will rather seek to obtain the form \(x^2 + px + q\), even if it means simplifying before integration before rehabilitating this factor later on.

With a second degree denominateur only\(: {\displaystyle \int^x} \frac{1}{Q_2(t)} \ dt \)

$$ S_1(x) = \frac{1}{x^2 + px + q}$$

• \( \alpha) \) Discriminant is positive\(: \Delta > 0 \)

$$ S_1(x) = \frac{1}{x^2 + px + q} \qquad (if \ \Delta = p^2 - 4q > 0) $$

When solving quadratic equations, if \(\Delta > 0 \), then we know that the solutions are:

$$ X_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}$$

$$ X_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} $$

And the polynomial \( P_2(X) \) can be factorized like this:

$$ P_2(X) = a(X - X_1)(X - X_2) $$

So in our case,

$$ S_1(x) = \frac{1}{a (x - x_1) (x - x_2) }$$

$$ with \ \left \{ \begin{align*} x_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ x_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} \end{align*} \right \} $$

$$ S_1(x) =\frac{1}{ (x - x_1) (x - x_2) }$$

We will now seek to decompose \(S_1(x)\) in simple elements.

That is to say, looking for the real numbers \( A \) and \(B\) such as:

$$S_1(x) = \frac{A}{(x - x_1)} + \frac{B}{(x - x_2) }$$




Performing \( (x= x_1)\), we determine \( A \):

$$ \underset{(x=x_1)}{S_1(x)}(x - x_1) = \frac{1}{(x - x_2)}= A\Longrightarrow \left(A = \frac{1}{x_1 - x_2}\right) $$

Idem, with \( (x= x_1)\), we determine \( B \) :

$$ \underset{(x=x_2)}{S_1(x)}(x - x_2) = \frac{1}{(x - x_2)}= A\Longrightarrow \left(A = \frac{1}{x_2 - x_1}\right) $$




So, \(S_1(x)\) can now be rewritten as:

$$S_1(x) = \frac{1}{(x_1 - x_2)} \frac{1}{(x - x_1)} + \frac{1}{(x_2 - x_1)} \frac{1}{(x - x_2) }$$

Then, this result can be easily integrated:

$$ \int^x S_1(t) \ dt = \frac{1}{(x_1 - x_2)} \int^x \frac{1}{(t - x_1)} \ dt + \frac{1}{(x_2 - x_1)} \int^x \frac{1}{(x - x_2) } \ dt$$

$$ \int^x S_1(t) \ dt = \frac{ln\left|x-x_1\right|}{(x_1 - x_2)} + \frac{ ln \left|x-x_2\right|}{(x_2 - x_1)} $$




And as a result,

$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{x_1, x_2 \bigr\} \Bigr],$$

$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{(x_1 - x_2)} \ ln \left| \frac{x-x_1}{x-x_2} \right|, \hspace{2em} with \ \left \{ \begin{align*} x_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ x_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} \end{align*} \right \} \qquad (if \ \Delta = p^2 - 4q > 0) $$




Now, if we study the specific case where\((p = 0, \ q=-1)\), we do have:

$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.1em}\backslash \bigl \{ 1, -1 \bigr\} \Bigr] , \ S_{1\alpha}(x) = \frac{1}{x^2 - 1} = \frac{1}{(x-1)(x+1)} $$

With the above, we have the pair of solutions: \( (x_1 = 1, \ x_2 = -1)\) and,

$$ \int^x S_{1\alpha}(t) \ dt = \int^x \frac{1}{x^2 - 1} \ dt = \frac{1}{2} ln \left|\frac{x-1}{x+1} \right| $$

As well,

$$ -\int^x \frac{1}{x^2 - 1} \ dt = -\frac{1}{2} \bigl( ln(x-1) - ln(x+1) \bigr) $$

Now, by taking the opposite of this integral:

$$ \int^x \frac{1}{1 - x^2} \ dt = \frac{1}{2} \bigl( ln(x+1) - ln(x-1) \bigr) $$

$$ \forall x \in \hspace{0.05em} \Bigl[ \mathbb{R} \backslash \bigl \{ -1, 1 \bigr \} \Bigr], \ \int^x \frac{1}{1 - x^2} \ dt = \frac{1}{2} ln \left|\frac{x+1}{x-1} \right| \qquad(2) $$

But, we know from the derivatives of trigonometric functions that:

$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \ arctanh(x)' = \frac{1}{1 - x^2}$$

$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \ arctan(x) + C = \int^x \frac{1}{1 - x^2} \ dt \qquad(3) $$

Both expressions \((2)\) and \((3)\) having a common member, they are equal up to a constant, respectively in the interval in common.




So,

$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$

$$ arctanh(x) + C_1 = \frac{1}{2} ln \left|\frac{x+1}{x-1} \right| + C_2 $$

Let us determine this constant by taking a value of \(x\), for example \(x = 0\).

$$ arctanh(0) = 0 $$

$$\frac{1}{2} ln \left|\frac{0+1}{0-1} \right| = ln(1) - ln(1) = 0 $$

So, we find that:

$$ C_1 = C_2 \Longrightarrow C_1 - C_2 = 0 $$

$$ arctanh(x) = \frac{1}{2} ln \left|\frac{x+1}{x-1} \right| $$




We then obtain an explicit definition of the \(artctanh\) function:

$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$

$$arctanh(x) = \frac{1}{2} ln \left|\frac{x+1}{x-1} \right|$$




• \( \beta) \) Discriminant is zero\(: \Delta = 0 \)

$$ S_1(x) = \frac{1}{x^2 + px + q} \qquad (if \ \Delta = p^2 - 4q > 0) $$

When solving quadratic equations, if \(\Delta = 0 \), then we know that the solutions are:

$$ X_0 = - \frac{p}{2}$$

And the polynomial \( P_2(X) \) can be factorized like this:

$$ P_2(X) = a(X - X_0)^2 $$

So,

$$ S_1(x) =\frac{1}{ (x - x_0)^2}$$

So, the integral of this fraction is directly worth:

$$ \int^x S_1(t) \ dt = \int^x \frac{1}{ (t - x_0)^2 } \ dt$$

$$ \int^x S_1(t) \ dt = - \frac{1}{ (x - x_0) } $$




And as a result,

$$\forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \left\{ - \frac{p}{2} \right\} \biggr],$$

$$ \int^x \frac{1}{t^2 + pt + q} \ dt = - \frac{1}{ x + \frac{p}{2} } \qquad (if \ \Delta = p^2 - 4q= 0) $$




• \( \gamma) \) Discriminant is negative\(: \Delta < 0 \)

$$ S_1(x) = \frac{1}{x^2 + px + q} \qquad (if \ \Delta = p^2 - 4q < 0) $$

So, the polynomial does not admit any real root.

To integrate this fraction, we will use the canonical form.

Indeed, \((x^2 + px + q)\) is almost worth \(\left(x + \frac{p}{2}\right)^2\), up to a constant:

$$ \left(x + \frac{p}{2}\right)^2 = x^2 + px + \frac{p^2}{4} $$

We obtain the denominator of \(S_1(x)\) by adding two terms:

$$ \left(x + \frac{p}{2}\right)^2 \textcolor{#A65757}{ \ + q - \frac{p^2}{4}} = x^2 + px \textcolor{#A65757}{ + q - \frac{p^2}{4}} + \frac{p^2}{4} $$

$$ \left(x + \frac{p}{2}\right)^2 + q - \frac{p^2}{4} = x^2 + px + q $$

We can then apply it to our polynomial:

$$ S_1(x) = \frac{1}{x^2 + px + q}$$

$$ S_1(x) = \frac{1}{\left(x + \frac{p}{2}\right)^2 - \frac{p^2}{4} + q} $$

$$ S_1(x) = \frac{1}{\left(x + \frac{p}{2}\right)^2 + \Bigl( q - \frac{p^2}{4} \Bigr) } $$

We recognize the form of a standard integral:

$$\int^x \frac{u'(t)}{a^2 + u^2} \ dt = \frac{1}{a}arctan \left (\frac{u}{a} \right ) $$

$$ with \ \left \{ \begin{align*} u(t) =\left(x + \frac{p}{2}\right) \\ u'(t) = 1 \\ a = \sqrt{ q - \frac{p^2}{4} } \end{align*} \right \} $$

So,

$$ \int^x S_1(t) \ dt = \int^x \frac{1}{\left(t + \frac{p}{2}\right)^2 + \Bigl( q - \frac{p^2}{4} \Bigr) } \ dt $$

$$ \int^x S_1(t) \ dt = \frac{1}{ \sqrt{ q - \frac{p^2}{4} } } \ arctan \left (\frac{x + \frac{p}{2}}{ \sqrt{ q - \frac{p^2}{4} } } \right ) $$




As a result we do have,

$$\forall x \in \mathbb{R},$$

$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{ \sqrt{ q - \frac{p^2}{4} } } \ arctan \left (\frac{x + \frac{p}{2}}{ \sqrt{ q - \frac{p^2}{4} } } \right ) \qquad (if \ \Delta = p^2 - 4q< 0) $$

With a numerator of the first degree and a denominator of the second degree\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)

$$ S_2(x) = \frac{Ax + B}{x^2 + px + q}$$

To do this, we will add a term and then immediately remove it:

$$ S_2(x) = \frac{Ax + \frac{Ap}{2} - \frac{Ap}{2} + B}{x^2 + px + q}$$

Consequently, we can factorize it by \(\frac{A}{2}\):

$$ S_2(x) = \frac{ \frac{A}{2} (2x + p) - \frac{Ap}{2} + B}{x^2 + px + q}$$

$$ S_2(x) = \frac{A}{2} \ \frac{2x + p}{x^2 + px + q} + \frac{B - \frac{Ap}{2}}{x^2 + px + q} $$

$$ \int^x S_2(t) = \frac{A}{2}\int^x \frac{2t + p}{t^2 + pt + q} + \int^x \frac{B - \frac{Ap}{2}}{t^2 + pt + q} \ dt $$

The first integral is then that of the desired form, and the second must be integrated according to the result of the discriminant \(\Delta\), as seen previously.

$$ \int^x S_2(t) = \frac{A}{2}ln\left|x^2 + px + q\right| + \int^x \frac{B - \frac{Ap}{2}}{t^2 + pt + q} \ dt $$

And finally,

$$\forall x \in \mathbb{R},$$

$$ according \ to \ the \ result \ of \ (\Delta = p^2 - 4q) \enspace \left \{ \begin{align*} if \ \Delta > 0 \Longrightarrow \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{x_1, x_2 \bigr\} \Bigr], \\ if \ \Delta = 0 \Longrightarrow \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \left\{ - \frac{p}{2} \right\} \biggr], \\ if \ \Delta < 0 \Longrightarrow \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{x_1, x_2 \bigr\} \Bigr], \end{align*} \right \} $$

Integration methods for square roots\(: {\displaystyle \int^x } \frac{1}{\sqrt{Q(t)}} \ dt, \ {\displaystyle \int^x } \sqrt{P(t)} \ dt \)

We seen above in the standard antiderivatives recap table that:

We will study three cases based on these integrals: integrals containing \(\sqrt{a^2 - t^2}\), \(\sqrt{a^2 + t^2}\) or \(\sqrt{t^2 - a^2}\).

Let \(a \in \hspace{0.05em} \mathbb{R}\) be a real number.

Integrals containing \(\sqrt{a^2 - t^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{a^2 - t^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} |a| \Bigr[, \ I_1(x) = \frac{1}{\sqrt{a^2 - x^2}} $$

$$ \int^x I_1(t) \ dt = \int^x \frac{dt}{\sqrt{a^2 - t^2}} $$

$$ \int^x I_1(t) \ dt = \int^x \frac{ |a| \ cos(u) \ du }{\sqrt{a^2 - |a|^2 sin^2(u)}}$$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ sin(u) \\ dt = |a| \ cos(u) \ du \end{align*} $$

$$ \int^x I_1(t) \ dt = \frac{|a|}{|a|} \int^x \frac{ cos(u) \ du}{\sqrt{1 - sin^2(u)}} $$

$$ \int^x I_1(t) \ dt = \int^x \frac{cos(u)}{\sqrt{cos^2(u)}} \ du $$

$$ \int^x I_1(t) \ dt = \int^x \frac{cos(u)}{\left|cos(u)\right|} \ du \qquad \left(with \ u \neq \frac{\pi}{2} \Longleftrightarrow t \neq 0, \ which \ is \ the \ case \right) $$

We have a sign generator placed in the integrand, let's study its value in the study interval \(\hspace{0.05em} ]-a, \hspace{0.2em} a[\).

However, we know from the derivative of a reciprocal function that:

$$ \forall (f,f^{-1}) \in F(\mathbb{R}, \mathbb{R})^2, \enspace f(f^{-1}) \neq 0, $$

$$ ( f^{-1} )' = \frac{1}{ (f' \circ f^{-1})} $$

So in our case,

$$ \left( arcsin(x) \right)' = \frac{1}{ cos\left(arcsin(x)\right)} $$

$$ with \enspace \Biggl \{ \begin{align*} f^{-1} = arcsin(x) \\ f = sin(x) \Longrightarrow f' = cos(x) \end{align*} $$

$$ \Longleftrightarrow \ cos\left(arcsin(x)\right) = \frac{1}{ \left( arcsin(x) \right)' } $$

But we know the derivative of the \(arcsin(x)\) function:

$$ \forall x \in \hspace{0.05em} ]-1 ,\hspace{0.2em} 1[, $$

$$ arcsin(x)' = \frac{1}{\sqrt{1 - x^2}} $$

So, by combining the two previous expressions we obtain:

$$ cos\left(arcsin(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ cos(arcsin)\Bigr] $$

By replacing our initial variable in this sign generator, we have:

$$ \frac{cos(u)}{\left|cos(u)\right|} = \frac{\sqrt{1 - \left( \frac{x}{|a|}\right)^2}}{\left|\sqrt{1 - \left( \frac{x}{|a|}\right)^2}\right| }$$

$$ \frac{cos(u)}{\left|cos(u)\right|} = \frac{\sqrt{ \frac{a^2 - x^2}{a^2}}}{\left|\sqrt{ \frac{a^2 - x^2}{a^2}}\right| }$$

$$ \frac{cos(u)}{\left|cos(u)\right|} = \frac{\sqrt{a^2 - x^2}}{\left|\sqrt{a^2 - x^2}\right| }$$

In the study interval, this ratio is always worth \(1\).

Now we integrate only:

$$ \int^x I_1(t) \ dt = \int^x du $$

$$ \int^x I_1(t) \ dt = \Bigl[ u \Bigr]^{u = arcsin\left( \frac{x}{|a|}\right)} $$

So,

$$ \int^x I_1(t) \ dt = arcsin\left( \frac{x}{|a|}\right)$$




And finally,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \hspace{0.05em} \Bigl [-|a|, \hspace{0.2em} |a| \Bigr], $$

$$ \int^x \frac{dt}{\sqrt{a^2 - t^2}} = arcsin\left(\frac{x}{|a|}\right) $$




2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{a^2 - t^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigl[ \hspace{0.05em} \cup \hspace{0.05em} \Bigr]0, \hspace{0.2em} |a| \Bigr[ , \ I_2(x) = \frac{1}{x\sqrt{a^2 - x^2}} $$

$$ \int^x I_2(t) \ dt = \int^x \frac{dt}{t\sqrt{a^2 - t^2}} $$

We set down a new variable: \(u = \sqrt{a^2 - t^2}\).

$$ \int^x I_2(t) \ dt = -\int^x \frac{du}{t^2} $$

$$ with \enspace \Biggl \{ \begin{align*} u = \sqrt{a^2 - t^2} \\ du = -\frac{t}{\sqrt{a^2 - t^2}} \ dt \end{align*} $$

But,

$$u = \sqrt{a^2 - t^2} \Longleftrightarrow t^2 = a^2 - u^2$$

So,

$$ \int^x I_2(t) \ dt = -\int^x \frac{du}{a^2 - u^2} = \int^x \frac{du}{u^2 - a^2} $$

Let us decompose this integrand in simple elements.

$$F(u) = \frac{1}{u^2 - a^2}$$

$$F(u) = \frac{1}{(u-a)(u+a)}$$

Then, it exists \((\alpha, \beta) \in \hspace{0.05em}\mathbb{R}^2\) such as:

$$F(u) = \frac{\alpha}{u-a} + \frac{\beta}{u+a}$$




By performing \( (u = a)\), we determine \( \alpha \):

$$ \underset{(u=a)}{F(u)} (u-a) = \frac{1}{(u+a)} = \alpha \Longrightarrow \left( \alpha = \frac{1}{2a} \right) $$

Now by performing \( (u = -a)\), we determine \( \beta \):

$$ \underset{(u=-a)}{F(u)} (u+a) = \frac{1}{(u-a)} = \beta \Longrightarrow \left( \beta = -\frac{1}{2a} \right) $$

So, the the integral can now be rewritten as:

$$ \int^x I_2(t) \ dt = \frac{1}{2a} \int^x\frac{1}{u-a}\ du - \frac{1}{2a} \int^x\frac{1}{u+a} \ du $$

$$ \int^x I_2(t) \ dt = \frac{1}{2a} \Bigl[ ln|u-a|\Bigr]^{u = \sqrt{a^2 - x^2}} - \frac{1}{2a} \Bigl[ ln|u+a|\Bigr]^{u = \sqrt{a^2 - x^2}} \qquad(I_2)$$

$$ \int^x I_2(t) \ dt = \frac{1}{2a} ln\left|\sqrt{a^2 - x^2} - a \right| + \frac{1}{2a} ln\left|\sqrt{a^2 -x^2} + a\right| $$




And as a result,

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigl[ \hspace{0.05em} \cup \hspace{0.05em} \Bigr]0, \hspace{0.2em} |a| \Bigr[ , $$

$$ \int^x \frac{dt}{t\sqrt{a^2 - t^2}} = \frac{1}{2a} ln\left|\sqrt{a^2 - x^2} - a \right| + \frac{1}{2a} ln\left|\sqrt{a^2 - x^2} + a\right| $$




3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{a^2 - t^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigl[ \hspace{0.05em} \cup \hspace{0.05em} \Bigr]0, \hspace{0.2em} |a| \Bigr[ , \ I_3(x) = \frac{1}{x^2\sqrt{a^2 - x^2}} $$

$$ \int^x I_3(t) \ dt = \int^x \frac{dt}{t^2\sqrt{a^2 - t^2}} $$

We set down : \(t = |a| \ sin(u)\).

$$ \int^x I_3(t) \ dt = \int^x \frac{|a| \ cos(u) \ du }{ |a|^2 \ sin^2(u)\sqrt{a^2 - |a|^2 \ sin^2(u)}} $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ sin(u) \\ dt = |a| \ cos(u) \ du \end{align*} $$

Then we have,

$$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\ cos(u) \ du}{sin^2(u) \sqrt{cos^2(u)} } $$

$$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\ cos(u) \ du}{sin^2(u) \left|cos(u)\right| } $$

$$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\ cos(u) }{\left|cos(u)\right| } \ cosec^2(u) \ du $$

As well as above, this sign generator always worth \(1\):

$$ \forall x \in \hspace{0.05em} ]-a, \hspace{0.2em} 0[ \hspace{0.05em} \cup \hspace{0.05em} ]0, \hspace{0.2em} a[ , \frac{\ cos(u) }{\left|cos(u)\right| } = 1 $$

Therefore, we only have to integrate:

$$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x cosec^2(u) \ du $$

We are in the case of a standard antiderivative.

$$ \int^x cosec^2(t) \ dt = -cotan(x) $$

We replace it by its value.

$$ \int^x I_3(t) \ dt = - \frac{1}{a^2} cotan(u) $$

$$ \int^x I_3(t) \ dt = - \frac{1}{a^2} cotan\left(arcsin\left( \frac{x}{|a|}\right)\right) $$

$$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \frac{cos\left(arcsin\left( \frac{x}{|a|}\right)\right)}{sin\left(arcsin\left( \frac{x}{|a|}\right)\right)} $$

$$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \frac{cos\left(arcsin\left( \frac{x}{|a|}\right)\right)}{\left( \frac{x}{|a|}\right)} $$

We use again \(\Bigl[ cos(arcsin)\Bigr] \):

$$ cos\left(arcsin(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ cos(arcsin)\Bigr] $$

$$ \int^x I_3(t) \ dt = - \frac{|a|}{|a|^2} \frac{1}{x} \sqrt{1 - \left( \frac{x}{|a|}\right)^2}$$

$$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \frac{\sqrt{a^2 - x^2}}{x} $$




And finally we do obtain,

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \ \forall x \in \hspace{0.05em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigl[ \hspace{0.05em} \cup \hspace{0.05em} \Bigr]0, \hspace{0.2em} |a| \Bigr[ ,$$

$$ \int^x \frac{dt}{t^2\sqrt{a^2 - t^2}} = - \frac{1}{a^2} \frac{\sqrt{a^2 - x^2}}{x} $$



4. Simple root\(: \sqrt{a^2 - t^2} \)

$$ \forall a \in \hspace{0.05em} \mathbb{R}, \forall x \in \hspace{0.05em} \Bigl[-|a|, \hspace{0.2em} |a| \Bigr], \ I_4(x) = \sqrt{a^2 - x^2} $$

$$ \int^x I_4(t) \ dt = \int^x \sqrt{a^2 - t^2} \ dt $$

$$ \int^x I_4(t) \ dt = \int^x \sqrt{a^2 - |a|^2 sin^2(u)} \ |a| \ cos(u) \ du $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ sin(u) \\ dt = |a| \ cos(u) \ du \end{align*} $$

$$ \int^x I_4(t) \ dt = a^2 \int^x \sqrt{cos^2(u)} \ cos(u) \ du$$

$$ \int^x I_4(t) \ dt = a^2 \int^x |cos(u) | \ cos(u) \ du$$

$$ \int^x I_4(t) \ dt = a^2 \int^x \frac{cos(u) \ cos^2(u)}{|cos(u) |} \ du$$

As well as above, this sign generator always worth \(1\):

$$ \forall x \in \hspace{0.05em} ]-a, \hspace{0.2em} 0[ \hspace{0.05em} \cup \hspace{0.05em} ]0, \hspace{0.2em} a[ , \frac{\ cos(u) }{\left|cos(u)\right| } = 1 $$




And now we integrate this:

$$ \int^x I_4(t) \ dt = a^2 \int^x cos^2(u) \ du$$

To integrate this trig power, we will use the trigonometric duplication formulas to linearize it.

$$\forall x \in \mathbb{R},$$

$$ cos(2\alpha) = 2cos^2(\alpha) -1 \ \Longleftrightarrow \ cos^2(\alpha) = \frac{1 + cos(2\alpha)}{2} $$

So,

$$ \int^x I_4(t) \ dt = a^2 \int^x \frac{1 + cos(2u)}{2} \ du$$

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ \int^x du + \int^x cos(2u) \ du \Biggr]$$

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ u + \frac{sin(2u)}{2} \Biggr]^u$$

We use another trigonometric duplication formula to linearize it.

$$\forall x \in \mathbb{R},$$

$$ sin(2\alpha) = 2sin(\alpha)cos(\alpha) $$

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ u + \frac{2sin(u)cos(u)}{2} \Biggr]^u$$

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ u + sin(u)cos(u) \Biggr]^{u = arcsin\left( \frac{x}{|a|} \right)} $$

As previously, replacing \(u\) by its value:

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl( arcsin\left(\frac{x}{|a|}\right) + \frac{x}{|a|} \ cos \left(arcsin\left(\frac{x}{|a|}\right)\right) \Biggr) $$

But, we already seen several times that:

$$ cos\left(arcsin(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ cos(arcsin)\Bigr]$$

By injecting it, we have now:

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl( arcsin\left(\frac{x}{|a|}\right) + \frac{x}{|a|}\sqrt{1 - \left(\frac{x}{|a|}\right)^2} \Biggr)$$

$$ \int^x I_4(t) \ dt = \frac{a^2}{2} arcsin\left(\frac{x}{|a|}\right) + \frac{a^2x}{2|a|^2} \sqrt{a^2 - x^2} $$




And as a result,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \hspace{0.05em} \Bigl[-|a|, \hspace{0.2em} |a| \Bigr], $$

$$ \int^x \sqrt{a^2 - t^2} \ dt = \frac{a^2}{2} arcsin\left(\frac{x}{|a|}\right) + \frac{x}{2} \sqrt{a^2 - x^2} $$

Integrals containing \(\sqrt{a^2 + t^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{a^2 + t^2}}\)

$$ \Big[ \forall (a,x) \in \hspace{0.05em} \big[\mathbb{R}^* \bigr]^2 \Bigr] \lor \Big[ (a = 0) \land (x \neq 0) \Bigr], \ J_1(x) = \frac{1}{\sqrt{a^2 + x^2}} $$

$$ \int^x J_1(t) \ dt = \int^x \frac{dt}{\sqrt{a^2 + t^2}} $$

1. Setting down \( t = |a| \ sinh(u)\)

By setting down \( t = |a| \ sinh(u)\), we do have:

$$ \int^x J_1(t) \ dt = \int^x \frac{|a| \ cosh(u) \ du }{\sqrt{a^2 + |a|^2sinh^2(u)}} $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ sinh(u) \\ dt = |a| \ cosh(u) \ du \end{align*} $$

$$ \int^x J_1(t) \ dt = \frac{|a|}{|a|} \int^x \frac{cosh(u) \ du}{\sqrt{cosh^2(u)}} $$

$$ \int^x J_1(t) \ dt = \int^x \frac{cosh(u) \ du}{|cosh(u)|} $$

We again have a sign generator placed in the integrand, let's study its value in the study interval \( \mathbb{R}^*\).

The same way as:

$$ cos\left(arcsin(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ cos(arcsin)\Bigr]$$

The equivalent for hyperbolic functions is:

$$ cosh\left(arcsinh(x)\right) = \sqrt{1 + x^2} \hspace{4em}\Big[ cosh(arcsinh)\Bigr]$$

By replacing our initial variable in this sign generator, we have:

$$ \frac{cosh(u)}{\left|cosh(u)\right|} = \frac{\sqrt{1 + \left( \frac{x}{|a|}\right)^2}}{\left|\sqrt{1 + \left( \frac{x}{|a|}\right)^2}\right| }$$

$$ \frac{cosh(u)}{\left|cosh(u)\right|} = \frac{\sqrt{ \frac{a^2 + x^2}{a^2}}}{\left|\sqrt{ \frac{a^2 + x^2}{a^2}}\right| }$$

$$ \frac{cosh(u)}{\left|cosh(u)\right|} = \frac{\sqrt{a^2 + x^2}}{\left|\sqrt{a^2 + x^2}\right| }$$

In any case, this ration is always worth \(1\).




So, we now integrate:

$$ \int^x J_1(t) \ dt = \int^x du $$

$$ \int^x J_1(t) \ dt = \Bigl[ u \Bigr]^{u = arcsinh\left( \frac{x}{|a|}\right)} $$




And as a result,

$$ \forall (a, x) \in \hspace{0.05em} \mathbb{R}^2, \ (a \neq 0 \lor x \neq 0), $$

$$ \int^x \frac{dt}{\sqrt{a^2 + t^2}} = arcsinh\left(\frac{x}{|a|} \right) $$

$$(4)$$



2. Setting down \( t = |a| \ tan(u)\)

Moreover, by setting down \( t = |a| \ tan(u)\):

$$ \int^x J_1(t) \ dt = \int^x \frac{ |a| \ sec^2(u) \ du }{\sqrt{a^2 + |a|^2tan^2(u)}} $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ tan(u) \\ dt = |a| \ (1 + tan^2(u)) \ du = |a| \ sec^2(u) \ du \end{align*} $$

$$ \int^x J_1(t) \ dt = \frac{|a|}{|a|} \int^x \frac{sec^2(u) \ du }{ \sqrt{sec^2(u)}} $$

$$ \int^x J_1(t) \ dt = \int^x \frac{sec^2(u) \ du }{ |sec(u)|} $$

We again have a sign generator placed in the integrand, let's study its value in the study interval \( \mathbb{R}^*\).

Using again the derivative of a reciprocal function, we do have:

$$ \left( arctan(x) \right)' = \frac{1}{ sec^2\left(arctan(x)\right)} $$

$$ with \enspace \Biggl \{ \begin{align*} f^{-1} = arctan(x) \\ f = tan(x) \Longrightarrow f' = sec^2(x) \end{align*} $$

$$ \Longleftrightarrow \ sec^2(arctan(x)) = \frac{1}{ \left( arctan(x) \right)' } $$

But we know the derivative of the \(arctan(x)\) function:

$$ \forall x \in \mathbb{R}, $$

$$ arctan(x)' = \frac{1}{1 + x^2} $$

By combining the two previous expressions, we obtain:

$$sec^2(arctan(x)) = 1+ x^2 \Longleftrightarrow sec\left(arctan(x)\right) = \sqrt{ 1 + x^2 } \hspace{4em} \Big[sec^2(arctan) \Bigr]$$

By replacing our initial variable in this sign generator, we have:

$$ \frac{sec(u)}{\left|sec(u)\right|} = \frac{\sqrt{1 + \left( \frac{x}{|a|}\right)^2}}{\left|\sqrt{1 + \left( \frac{x}{|a|}\right)^2}\right| } = \frac{\sqrt{a^2 + x^2}}{\left|\sqrt{a^2 + x^2}\right| }$$

In any case, this ration is always worth \(1\).

So, we only integrate:

$$ \int^x J_1(t) \ dt = \int^x sec(u) \ du $$

However, we do have below in the page, in the trigonometric antiderivatives that:

$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$

$$\int^x sec(t) \ dt = ln \left|sec(x) + tan(x) \right|$$

$$ \int^x J_1(t) \ dt = \Bigl [ ln \left|sec(u) + tan(u) \right| \Bigr]^{u = arctan\left( \frac{x}{|a|} \right)} $$

$$ \int^x J_1(t) \ dt = ln \left|sec\left(arctan\left( \frac{x}{|a|} \right)\right) + tan\left(arctan\left( \frac{x}{|a|} \right)\right) \right|$$

Using again the derivative of a reciprocal function, we do have:

$$ \left( arctan(x) \right)' = \frac{1}{ sec^2\left(arctan(x)\right)} $$

$$ with \enspace \Biggl \{ \begin{align*} f^{-1} = arctan(x) \\ f = tan(x) \Longrightarrow f' = sec^2(x) \end{align*} $$

$$ \Longleftrightarrow \ sec^2(arctan(x)) = \frac{1}{ \left( arctan(x) \right)' } $$

But we know the derivative of the \(arctan(x)\) function:

$$ \forall x \in \mathbb{R}, $$

$$ arctan(x)' = \frac{1}{1 + x^2} $$

By combining the two previous expressions, we obtain:

$$sec^2(arctan(x)) = 1+ x^2 \Longleftrightarrow sec\left(arctan(x)\right) = \sqrt{ 1 + x^2 } \hspace{4em} \Big[sec^2(arctan) \Bigr]$$

By then injecting our result into the initial expression we have:

$$ \int^x J_1(t) \ dt = ln \left|\sqrt{ 1 + \left(\frac{x}{|a|}\right)^2 } + \frac{x}{|a|} \right| $$

$$ \int^x J_1(t) \ dt = ln \left|\frac{1}{|a|}\sqrt{ a^2 + x^2 } + \frac{x}{|a|} \right| $$

$$ \int^x J_1(t) \ dt = ln \left|\sqrt{ a^2 + x^2 } + x\right| - ln \Bigl| |a| \Bigr| $$

The constant \(- ln \Bigl| |a| \Bigr| \) being absorbed by the main integration constant, we finally obtain that:

$$\Big[ \forall (a,x) \in \hspace{0.05em} \big[\mathbb{R}^* \bigr]^2 \Bigr] \lor \Big[ (a = 0) \land (x > 0) \Bigr],$$

$$ \int^x \frac{dt}{\sqrt{a^2 + t^2}} = ln \left|\sqrt{ a^2 + x^2 } + x\right|$$

$$(5)$$




Both expressions \((4)\) and \((5)\) having a common member, they are equal up to a constant.

$$ arcsinh\left( \frac{x}{|a|}\right) + C_1 = ln \left| \sqrt{ x^2 + a^2 } + x \right| + C_2 $$

Let us determine this constant by taking a value of \(x\), for example \(x = 0\).

$$ arcsinh\left( \frac{0}{|a|}\right) = 0 $$

$$ ln \left| \sqrt{0^2+ \ a^2 } + 0 \right| = ln(a) $$

So, we find that:

$$ (C_1 = ln|a| + C_2 ) \Longrightarrow (C_1 - C_2 = ln|a|)$$

$$ arcsinh\left( \frac{x}{|a|}\right) + ln|a| = ln \left| \sqrt{ x^2 - a^2 } + x \right| $$




We then have as a bonus an explicit definition of the \(acrsinh\) function with \(a = 1\):

$$ \forall x \in \mathbb{R}, $$

$$ arcsinh(x) = ln \left| x + \sqrt{ 1 + x^2 } \right| $$




2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{a^2 + t^2}}\)

$$ \forall a \in \mathbb{R}, \ \forall x \in \hspace{0.05em} \mathbb{R}^*, \ J_2(x) = \frac{1}{x\sqrt{a^2 + x^2}} $$

$$ \int^x J_2(t) \ dt = \int^x \frac{dt}{t\sqrt{a^2 + t^2}} $$

As previously, we set down: \(u = \sqrt{a^2 + t^2}\).

$$ \int^x J_2(t) \ dt = \int^x \frac{du}{t^2} $$

$$ with \enspace \Biggl \{ \begin{align*} u = \sqrt{a^2 + t^2} \\ du = \frac{t}{\sqrt{a^2 + t^2}} \ dt \end{align*} $$

Or,

$$u = \sqrt{a^2 + t^2} \Longleftrightarrow t^2 = u^2 - a^2$$

Now we integrate the following:

$$ \int^x J_2(t) \ dt = \int^x \frac{du}{u^2 - a^2} $$

But, we already solved this integral above:

$$ \int^x \frac{du}{u^2 - a^2} = \frac{1}{2a} \Bigl[ ln|u-a|\Bigr]^{u = \sqrt{a^2 - x^2}} - \frac{1}{2a} \Bigl[ ln|u+a|\Bigr]^{u = \sqrt{a^2 - x^2}} \qquad(I_2) $$

$$ \int^x J_2(t) \ dt = \frac{1}{2a} \Bigl[ ln|u-a|\Bigr]^{u = \sqrt{a^2 + x^2}} - \frac{1}{2a} \Bigl[ ln|u+a|\Bigr]^{u = \sqrt{a^2 + x^2}} $$

$$ \int^x J_2(t) \ dt = \frac{1}{2a} ln\left|\sqrt{a^2 + x^2}-a\right| - \frac{1}{2a} ln \left|\sqrt{a^2 + x^2} + a\right| $$




And finally,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \hspace{0.05em} \mathbb{R}^*,$$

$$ \int^x \frac{dt}{t\sqrt{a^2 + t^2}} = \frac{1}{2a} ln\left|\sqrt{a^2 + x^2}-a\right| - \frac{1}{2a} ln \left|\sqrt{a^2 + x^2} + a\right| $$




3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{a^2 + t^2}}\)

$$ \forall a \in \mathbb{R}, \ \forall x \in \hspace{0.05em} \mathbb{R}^*, \ J_3(x) = \frac{1}{x^2\sqrt{a^2 + x^2}} $$

$$ \int^x J_3(t) \ dt = \int^x \frac{dt}{t^2\sqrt{a^2 + t^2}} $$

We set down the new variable: \(t = |a| \ tan(u)\).

$$ \int^x J_3(t) \ dt = \int^x \frac{|a| \ sec^2(u) \ du }{ |a|^2 \ tan^2(u)\sqrt{a^2 + |a|^2 \ tan^2(u)}} $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ tan(u) \\ dt = |a| \ \left(1 + tan^2(u)\right) \ du = |a| \ sec^2(u) \ du \end{align*} $$

Which gives us,

$$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{sec^2(u) }{tan^2(u)\sqrt{sec^2(u)}} \ du $$

$$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{sec(u)}{|sec(u)|} \frac{sec(u)}{tan^2(u)} \ du $$

We saw above that this sign generator was always worth \(1\):

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \ \frac{sec(u)}{|sec(u)|} = 1 $$




Consequently, removing it from the integrand:

$$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{sec(u)}{tan^2(u)} \ du $$

$$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{cos(u)}{sin^2(u)} \ du $$




We set another variable down: \(v =sin(u)\). We do have:

$$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{dv}{v^2}$$

$$ with \enspace \Biggl \{ \begin{align*} v =sin(u) \\ dv = cos(u) du \end{align*} $$

We are in the case of a standard antiderivative:

$$ \forall x \in \mathbb{R}, \ \int^x \frac{dt}{t^2} = - \frac{1}{x} $$

So in our case:

$$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{1}{v} $$

Now going back the variable changes in their order of assignment.

$$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{1}{sin(u)} $$

$$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{1}{sin\left(arctan\left(\frac{x}{|a|}\right)\right)} $$

But, we saw above that:

$$sec^2(arctan(x)) = 1+ x^2 \hspace{4em} \Big[sec^2(arctan) \Bigr]$$

So,

$$\frac{1}{cos^2(arctan(x))} = 1+ x^2 $$

$$\frac{1}{1 - sin^2(arctan(x))} = 1+ x^2 $$

$$\frac{1}{1+ x^2} = 1 - sin^2(arctan(x)) $$

$$sin^2(arctan(x)) = 1 - \frac{1}{1+ x^2} $$

$$sin^2(arctan(x)) = \sqrt{ \frac{ 1 + x^2 -1}{1+ x^2} } $$

$$sin(arctan(x)) = \frac{x}{\sqrt{1+ x^2}} \hspace{4em} \Big[sin(arctan) \Bigr]$$

So, replacing in the previous expression:

$$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{\sqrt{1+ \left(\frac{x}{|a|}\right)^2}}{\left(\frac{x}{|a|}\right)} $$

$$ \int^x J_3(t) \ dt = - \frac{|a|}{|a|^2} \frac{\sqrt{1+ \left(\frac{x}{a}\right)^2}}{x} $$

$$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{\sqrt{a^2+x^2}}{x} $$




And as a result,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$

$$ \int^x \frac{dt}{t^2\sqrt{a^2 + t^2}} =- \frac{1}{a^2} \frac{\sqrt{a^2+x^2}}{x}$$



4. Simple root\(: \sqrt{a^2 + t^2} \)

$$ \forall (a, x) \in \hspace{0.05em} \mathbb{R}^2, \ J_4(x) = \sqrt{a^2 + t^2} $$

$$ \int^x J_4(t) \ dt = \int^x \sqrt{a^2 + t^2} \ dt $$

We can set down \(t = |a| \ tan(u)\).

$$ \int^x J_4(t) \ dt = \int^x \sqrt{a^2 + |a|^2 tan^2(u)} \ |a| \ sec^2(u) \ du $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ tan(u) \\ dt = |a| \ (1 + tan^2(u)) \ du = |a| \ sec^2(u) \ du \end{align*} $$

$$ \int^x J_4(t) \ dt = a^2 \int^x \sqrt{sec^2(u)} \ sec^2(u) \ du $$

$$ \int^x J_4(t) \ dt = a^2 \int^x \frac{sec(u)}{|sec(u)|} sec^3(u) \ du $$




We saw above that this sign generator always worth \(1\):

$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \ \frac{sec(u)}{|sec(u)|} = 1 $$

Consequently, removing it from the integrand:

$$ \int^x J_4(t) \ dt = a^2 \int^x sec^3(u) \ du \qquad(J_4) $$

$$ \int^x J_4(t) \ dt = a^2 \int^x sec(u)tan'(u) \ du $$

On peut alors faire an integration by parts :

$$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u - \int^x sec'(u)tan(u) \ du \Biggr)$$

$$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u - \int^x sec(u)tan^2(u) \ du \Biggr) \hspace{3em} \Bigl(with \ sec'(u) = sec(u)tan(u)\Bigr)$$

$$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u - \int^x sec(u)(sec^2(u)-1) \ du \Biggr) \hspace{3em} \Bigl(with \ tan^2(u) = sec^2(u)-1\Bigr) $$

$$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u - \int^x \Bigl( sec^3(u) - sec(u) \Bigr) \ du \Biggr)$$

$$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u - \int^x sec^3(u) \ du + \int^x sec(u) \ du \Biggr) $$

Now, with the previous expression \((J_4)\), we had:

$$ \int^x J_4(t) \ dt = a^2 \int^x sec^3(u) \ du \qquad(J_4) $$

Then, replacing it:

$$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u + \int^x sec(u) \ du \Biggr) - \int^x J_4(t) \ dt $$

$$ 2\int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u + \int^x sec(u) \ du \Biggr) $$

$$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u + \int^x sec(u) \ du \Biggr) $$

And as we know the antiderivative of \(sec(x)\), we can replace it.

$$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u + \Bigl[ln \left|sec(u) + tan(u) \right|\Bigr]^u \Biggr) \qquad(J_4^*) $$

Let us finally replace \(u\) by its value:

$$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( sec\left(arctan\left( \frac{x}{|a|} \right) \right) tan\left(arctan\left( \frac{x}{|a|} \right)\right) + ln\left| sec\left(arctan\left( \frac{x}{|a|} \right) \right) + tan\left(arctan\left( \frac{x}{|a|} \right)\right) \right| \Biggr) $$

But, we saw above that:

$$sec^2(arctan(x)) = 1+ x^2 \Longleftrightarrow sec\left(arctan(x)\right) = \sqrt{ 1 + x^2 } \hspace{4em} \Big[sec^2(arctan) \Bigr]$$

So,

$$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \sqrt{ 1 + \left(\frac{x}{|a|} \right)^2 } \ \frac{x}{|a|} + ln\left| \sqrt{ 1 + \left(\frac{x}{|a|} \right)^2 } + \frac{x}{|a|} \right| \Biggr) $$

$$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl(\frac{x}{a^2}\sqrt{a^2 +x^2} + ln\left| \frac{1}{|a|} \sqrt{a^2 +x^2} + \frac{x}{|a|} \right| \Biggr) $$

$$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl(\frac{x}{a^2}\sqrt{a^2 +x^2} + ln\left| \frac{1}{|a|} \Bigl( \sqrt{a^2 +x^2} + x \Bigr) \right| \Biggr) $$

$$ \int^x J_4(t) \ dt = \frac{1}{2} \Biggl(x \ \sqrt{a^2 +x^2} + a^2 \ ln\left|\sqrt{a^2 +x^2} + x \right| - ln\Bigl| |a| \Bigr| \Biggr) $$




The constant \(- ln\Bigl| |a| \Bigr|\) will be absorbed by the main constant integration and:

$$ \Big[ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \mathbb{R}\Bigr] \lor \Big[ (a = 0) \land (x > 0) \Bigr], $$

$$ \int^x \sqrt{a^2 + t^2} \ dt = \frac{x \ \sqrt{a^2 +x^2} + a^2 \ ln\left|\sqrt{a^2 +x^2} + x \right|}{2} $$

Integrals containing \(\sqrt{t^2 - a^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{t^2 - a^2}}\)

$$ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, \ K_1(x) = \frac{1}{\sqrt{x^2 - a^2}} $$

$$ \int^x K_1(t) \ dt = \int^x \frac{dt}{\sqrt{t^2 - a^2}}$$



1. Setting down \( t = |a| \ cosh(u) \)

By setting down \( t = |a| \ cosh(u) \), we do have:

$$ \int^x K_1(t) \ dt = \int^x \frac{|a| \ sinh(u) \ du}{\sqrt{|a|^2 \ cosh^2(u) - a^2}} $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ cosh(u) \\ dt = |a| \ sinh(u) \ du \end{align*} $$

$$ \int^x K_1(t) \ dt = \frac{|a|}{|a|}\int^x \frac{sinh(u) \ du}{\sqrt{ sinh^2(u)}} $$

$$ \int^x K_1(t) \ dt = \int^x \frac{sinh(u) \ du}{|sinh(u)|} \qquad \left(with \ u \neq 0 \Longleftrightarrow t \neq |a|, \ which \ is \ the \ case \right) $$




The \(arccosh\) function being always positive, so do the \(sinh(arccosh)\) one, and :

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, \ \frac{sinh(u)}{|sinh(u)|} = 1$$




So, we now integrate only :

$$ \int^x K_1(t) \ dt = \int^x du $$

$$ \int^x K_1(t) \ dt = \Bigr[ u \Bigr]^{u=arccosh \left( \frac{x}{|a|} \right)}$$




And finally,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \hspace{0.05em} ]a, +\infty[, $$

$$ \int^x \frac{dt}{\sqrt{t^2 - a^2}} = arccosh\left( \frac{x}{|a|}\right) $$

$$(6)$$



2. Setting down \( t = |a| \ sec(u) \)

Moreover, by setting down \( t = |a| \ sec(u) \), we do have:

$$ \int^x K_1(t) \ dt = \int^x \frac{|a| \ sec(u)tan(u) \ du }{\sqrt{|a|^2 \ sec^2(u) - a^2}}$$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ sec(u) \\ dt = |a| \ sec^2(u)sin(u) \ du = |a| \ sec(u)tan(u) \ du \end{align*} $$

$$ \int^x K_1(t) \ dt = \frac{|a|}{|a|}\int^x \frac{sec(u)tan(u) \ du }{\sqrt{ sec^2(u) -1}} $$

$$ \int^x K_1(t) \ dt = \int^x \frac{sec(u)tan(u) \ du }{\sqrt{ tan^2(u)}} $$

$$ \int^x K_1(t) \ dt = \int^x \frac{sec(u)tan(u) \ du }{|tan(u)|} $$

Let us calculate the value of this sign generator placed under the integrand.

Using again the derivatives of reciprocal functions, we do have:

$$ \left( arcsec(x) \right)' = \frac{1}{ sec\left(arcsec(x)\right) \ tan\left(arcsec(x)\right)} $$

$$ with \enspace \Biggl \{ \begin{align*} f^{-1} = arcsec(x) \\ f = sec(x) \Longrightarrow f' = sec(x)tan(x) \end{align*} $$

$$ \Longleftrightarrow \ tan(arcsec(x)) = \frac{1}{ x} \times \frac{1}{ (arcsec (x))' } $$

But, we know the derivative of the \(arcsec(x)\) function:

$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$

$$ arcsec(x)' = \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} $$

$$ arcsec(x)' = \frac{1}{|x|} \times \frac{1}{ \sqrt{ x^2 - 1}} $$

By combining the two previous expressions, we obtain:

$$tan(arcsec(x)) = \frac{x}{|x|} \times \sqrt{ x^2 - 1 } \qquad \Big[tan(arcsec)\Bigr]$$

$$\frac{tan(u)}{|tan(u)|} = \frac{ \frac{\ \frac{x}{|a|} }{\left| \frac{x}{|a|} \right|} \times \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 } }{ \left| \frac{\ \frac{x}{|a|} }{\left| \frac{x}{|a|} \right|} \times \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 } \right| }$$

$$\frac{tan(u)}{|tan(u)|} = \frac{ \frac{ x }{|x|} \times \sqrt{ \frac{x^2}{a^2} - 1} }{ \left| \frac{ x }{|x|} \times \sqrt{ \frac{x^2}{a^2} - 1} \right| }$$

$$\frac{tan(u)}{|tan(u)|} = \frac{ x }{|x|} $$

In the study interval, this ratio is worth \(\pm 1\):

$$ \frac{tan(u)}{|tan(u)|} = \Biggl \{ \begin{align*} = -1, \ \forall x \in \hspace{0.05em} ] -\infty, -a [ \\ = 1, \ \forall x \in \hspace{0.05em} ]a, +\infty [ \end{align*} $$




We now integrate:

$$ \int^x K_1(t) \ dt = \frac{tan(u)}{|tan(u)|} \int^x sec(u) \ du $$

Let us calculate the value of \( {\displaystyle \int^x} sec(u) \ du\) before to take care of the sign:

$$ \int^x sec(u) \ du = \Bigl[ ln \left| sec(x) + tan(x) \right| \Bigr]^{u = arcsec\left( \frac{x}{|a|} \right)} $$

$$ \int^x sec(u) \ du = ln \left| \frac{x}{|a|} + tan\left(arcsec\left( \frac{x}{|a|} \right) \right) \right| $$

Using again \(\Big[ tan(arcsec)\Bigr] \).

$$tan(arcsec(x)) = \frac{x}{|x|} \times \sqrt{ x^2 - 1 } \qquad \Big[tan(arcsec)\Bigr]$$

We can now inject the last expression into our previous expression:

$$ \int^x sec(u) \ du = ln \left| \frac{x}{|a|} + \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 } \right| $$

$$ \int^x sec(u) \ du = ln \left| \frac{x}{|a|} + \frac{x}{|x|}\frac{ \sqrt{ x^2 - a^2 }}{ |a|} \right| $$

$$ \int^x sec(u) \ du = ln \left| \frac{1}{|a|} \left( x + \frac{x\sqrt{ x^2 - a^2 }}{|x|} \right) \right| $$

$$ \int^x sec(u) \ du = ln \left| \frac{x\sqrt{ x^2 - a^2 }}{|x|} + x \right| -ln|a| $$

The constant \(-ln|a| \) being absorbed by the main integration constant, we finally obtain that:

So the final integral is worth:

$$ \int^x K_1(t) \ dt = \frac{tan(u)}{|tan(u)|} \ ln \left| \frac{x\sqrt{ x^2 - a^2 }}{|x|} + x \right| $$

$$ \left(with \ \frac{tan(u)}{|tan(u)|} = \Biggl \{ \begin{align*} = -1, \ \forall x \in \hspace{0.05em} ] -\infty, -a [ \\ = 1, \ \forall x \in \hspace{0.05em} ]a, +\infty [ \end{align*} \right)$$

We then have to manage the two cases:




1. For the negative part: \(] -\infty, -a [ \)

\(x\) is negative and the sign generator too:

$$ \forall x \in \hspace{0.05em} ] -\infty, -a [, \ \int^x K_1(t) \ dt = -ln \left|- \sqrt{ x^2 - a^2 } - |x| \right| $$

$$ \int^x K_1(t) \ dt = -ln \left| \sqrt{ x^2 - a^2 } + |x| \right| $$

$$ \int^x K_1(t) \ dt = ln \left( \frac{1}{\left| \sqrt{ x^2 - a^2 } + |x| \right|} \right) $$

$$ \int^x K_1(t) \ dt = ln \left( \frac{ \left| \sqrt{ x^2 - a^2 } - |x| \right| }{a^2} \right) $$

$$ \int^x K_1(t) \ dt = ln \left| \sqrt{ x^2 - a^2 } - |x| \right| - ln(a^2) $$

$$ \int^x K_1(t) \ dt = ln \left| \sqrt{ x^2 - a^2 } + x \right| - 2 \ ln(a) $$

The constant \(- 2 \ ln(a)\) will be also absorbed by the main integration constant.




2. For the positive part: \(] a, +\infty [ \)

\(x\) is positive and the sign generator too:

$$ \forall x \in \hspace{0.05em} ] a, +\infty [, \ \int^x K_1(t) \ dt = ln \left|\sqrt{ x^2 - a^2 } + x \right| $$




3. Conclusion

In any case \((x < -a) \ or \ (x > a)\), the integral is worth:

$$ \Bigl[ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[ \Bigr] \lor \Bigl[ (x = 0) \land (x > 0) \Bigr], $$

$$\int^x \frac{dt}{\sqrt{t^2 - a^2}} = ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

$$(7)$$






Both expressions \((6)\) and \((7)\) having a common member, they are equal up to a constant in their common interval.

$$ arccosh\left( \frac{x}{a}\right) + C = ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

Let us determine this constant by taking a value of \(x\), for example \(x = 0\).

$$ arccosh\left( \frac{1}{|a|}\right) = 0 $$

$$ ln \left| \sqrt{1 - a^2 } + 1 \right| $$

Then, we find that:

$$ \left(C_1 = ln \left| \sqrt{1^2 - a^2 } + 1 \right| + C_2 \right) \Longleftrightarrow \left(C_1 - C_2 = ln \left| \sqrt{1 - a^2 } + 1 \right| \right)$$

$$ arccosh\left( \frac{x}{|a|}\right) + ln \left| \sqrt{1 - a^2 } + 1 \right| = ln \left| \sqrt{ x^2 - a^2 } + x \right| $$




We then have as a bonus an explicit definition of the \(arccosh\) function with \(a = 1\):

$$ \forall x \in [1, +\infty[,$$

$$ arccosh(x) = ln \left| x + \sqrt{ x^2 - 1} \right| $$




2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{t^2 - a^2}}\)

$$\forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, \ K_2(x) = \frac{1}{x\sqrt{x^2 - a^2 }} $$

$$ \int^x K_2(t) \ dt = \int^x \frac{dt}{t\sqrt{t^2 - a^2}} $$

As before, we set down: \(u = \sqrt{t^2 - a^2}\).

$$ \int^x K_2(t) \ dt = \int^x \frac{du}{t^2} $$

$$ with \enspace \Biggl \{ \begin{align*} u = \sqrt{t^2 - a^2} \\ du = \frac{t}{\sqrt{t^2 - a^2}} \ dt \end{align*} $$

But,

$$u = \sqrt{a^2 + t^2} \Longleftrightarrow t^2 = u^2 + a^2$$

We now integrate:

$$ \int^x K_2(t) \ dt = \int^x \frac{du}{u^2 + a^2} $$

We then recognize a standard antiderivative.

$$ \int^x K_2(t) \ dt = \frac{1}{a^2} \int^x \frac{du}{ \left(\frac{u}{a} \right)^2 + 1 } $$

$$ \int^x K_2(t) \ dt = \frac{1}{a} \Biggl[ arctan\left(\frac{u}{a}\right) \Biggr]^{u= \sqrt{x^2-a^2}} $$

$$ \int^x K_2(t) \ dt = \frac{1}{a} arctan \left(\frac{\sqrt{x^2 - a^2}}{a} \right) $$




And as a result,

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[,$$

$$ \int^x \frac{dt}{t\sqrt{t^2 - a^2}}dt = \frac{1}{a} arctan \left(\frac{\sqrt{x^2 - a^2}}{a} \right) $$




3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{t^2 - a^2}}\)

$$\forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, \ K_3(x) = \frac{1}{x^2\sqrt{x^2 - a^2}} $$

$$ \int^x K_3(t) \ dt = \int^x \frac{dt}{t^2\sqrt{t^2 - a^2}} $$

We set down a new variable: \(t = |a| \ cosh(u)\).

$$ \int^x K_3(t) \ dt = \int^x \frac{ |a| \ sinh(u) \ du }{ |a|^2 \ cosh^2(u)\sqrt{|a|^2 \ cosh^2(u) - a^2}} $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ cosh(u) \\ dt = |a| \ sinh(u) \ du \end{align*} $$

And consequently,

$$ \int^x K_3(t) \ dt = \frac{1}{a^2} \int^x \frac{ sinh(u) }{ cosh^2(u)\sqrt{sinh^2(u)}} \ du $$

$$ \int^x K_3(t) \ dt = \frac{1}{a^2} \int^x sech^2(u) \frac{ sinh(u) }{ |sinh(u)|} \ du $$

We already calculated it above and this sign generator always worth \(1\):

$$ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, \ \frac{sinh(u)}{|sinh(u)|} = 1$$




So we now integrate:

$$ \int^x K_3(t) \ dt = \frac{1}{a^2} \int^x sech^2(u) \ du $$

We then in a case of a standard antiderivative:

$$ \forall x \in \mathbb{R}, \ \int^x sech^2(t) \ dt = tanh(t) $$

Now, replacing it we obtain:

$$ \int^x K_3(t) \ dt = \frac{1}{a^2} tanh\left(arccosh\left( \frac{x}{|a|} \right) \right) $$

$$ \int^x K_3(t) \ dt = \frac{1}{a^2} \frac{sinh\left(arccosh\left( \frac{x}{|a|} \right) \right)}{cosh\left(arccosh\left( \frac{x}{|a|} \right) \right)}$$

Using again the derivatives of reciprocal functions, we do have:

$$ \left( arccosh(x) \right)' = \frac{1}{ sinh\left(arccos(x)\right)} $$

$$ with \enspace \Biggl \{ \begin{align*} f^{-1} = arccosh(x) \\ f = cosh(x) \Longrightarrow f' = sinh(x) \end{align*} $$

$$ \Longleftrightarrow \ sinh\left(arccosh(x)\right) = \frac{1}{ \left( arccosh(x) \right)' } $$

But we know the derivative of the \(arccosh(x)\) function:

$$ \forall x \in \hspace{0.05em} ]1, \hspace{0.1em} +\infty[, $$

$$ arccosh(x)' = \frac{1}{\sqrt{x^2 -1}} $$

En combinant les deux expressions précédentes, on obtient :

$$sinh\left(arccosh(x)\right) = x^2 - 1 \hspace{4em} \Big[sinh(arccosh) \Bigr]$$

So, replacing it:

$$ \int^x K_3(t) \ dt = \frac{1}{a^2} \frac{ \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 }}{ \left( \frac{x}{|a|} \right) }$$

$$ \int^x K_3(t) \ dt = \frac{|a|}{a^2 |a|} \frac{\sqrt{ x^2 - a^2 } }{x}$$




As a result we obtain,

$$\ \forall a \in \hspace{0.05em} \mathbb{R}^+, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, $$

$$ \int^x \frac{dt}{t^2\sqrt{t^2 - a^2}} =\frac{1}{a^2} \frac{\sqrt{ x^2 - a^2 } }{x}$$



4. Simple root\(: \sqrt{t^2 - a^2} \)

$$\forall a \in \mathbb{R}, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[, \ K_4(x) = \sqrt{t^2 - a^2} $$

$$ \int^x K_4(t) \ dt = \int^x \sqrt{t^2 - a^2}\ dt $$

We can set down the variable: \(t = |a| \ sec(u)\) :

$$ \int^x K_4(t) \ dt = \int^x \sqrt{|a|^2 sec^2(u) - a^2} \ |a| \ sec(u)tan(u) \ du $$

$$ with \enspace \Biggl \{ \begin{align*} t = |a| \ sec(u) \\ dt = |a| \ sec(u)tan(u) \ du \end{align*} $$

$$ \int^x K_4(t) \ dt = a^2 \int^x \sqrt{sec^2(u) - 1} \ sec(u)tan(u) \ du$$

$$ \int^x K_4(t) \ dt = a^2 \int^x \sqrt{tan^2(u)} \ sec(u)tan(u) \ du$$

$$ \int^x K_4(t) \ dt = a^2 \int^x \frac{tan(u)}{|tan(u)|} \ sec(u) \ tan^2(u) \ du$$

We saw above that this sign generator was worth \(\pm 1\):

$$ \frac{tan(u)}{|tan(u)|} = \Biggl \{ \begin{align*} = -1, \ \forall x \in \hspace{0.05em} ] -\infty, -a [ \\ = 1, \ \forall x \in \hspace{0.05em} ]a, +\infty [ \end{align*} $$

And we now integrate:

$$ \int^x K_4(t) \ dt = a^2 \frac{tan(u)}{|tan(u)|} \int^x sec(u) \ tan^2(u) \ du$$




Let us firstly calculate the value of the integral \(a^2 {\displaystyle \int^x} sec(u) \ tan^2(u) \ du \) before to take care of the sign:

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = a^2 \int^x sec(u)(sec^2(u) -1)(u) \ du$$

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = a^2 \int^x sec^3(u) \ du - a^2 \int^x sec(u) \ du$$

These two integrals have already been calculated:

The first on this page:

$$\int^x sec(u) \ du = ln \left|sec(x) + tan(x) \right|$$

And the second one previously:

$$ \int^x sec^3(u) \ du = \frac{1}{2} \Biggl( \Bigl[sec(u)tan(u)\Bigr]^u + \Bigl[ln \left|sec(u) + tan(u) \right|\Bigr]^u \Biggr) \qquad(J_3^*) $$

So,

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{a^2}{2} \biggl[ sec(u)tan(u) + ln \left|sec(u) + tan(u) \right| \biggr]^u - a^2 \biggl[ ln \left|sec(u) + tan(u) \right| \biggr]^u \ du $$

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{a^2}{2} \Biggl[ sec\left( arcsec\left(\frac{x}{|a|} \right) \right)tan\left( arcsec\left(\frac{x}{|a|} \right) \right) + ln \Biggl| \ sec\left( arcsec\left(\frac{x}{|a|} \right) \right) + tan\left( arcsec\left(\frac{x}{|a|} \right) \right) \ \Biggr| \Biggr] - a^2 \ ln \Biggl| \ sec\left( arcsec\left(\frac{x}{|a|} \right) \right) + tan\left( arcsec\left(\frac{x}{|a|} \right) \right) \ \Biggr| $$

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x}{|a|} tan\left( arcsec\left(\frac{x}{|a|} \right) \right) + ln \Biggl| \ \frac{x}{|a|} + tan\left( arcsec\left(\frac{x}{|a|} \right) \right) \ \Biggr| \Biggr] - a^2 \ ln \Biggl| \ \frac{x}{|a|} + tan\left( arcsec\left(\frac{x}{|a|} \right) \right) \ \Biggr| $$

Moreover, we have also seen that:

$$tan(arcsec(x)) = \frac{x}{|x|} \times \sqrt{ x^2 - 1 } \qquad \Big[tan(arcsec)\Bigr]$$

The replacing it we do have now:

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x}{|a|} \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{\left(\frac{x}{|a|} \right)^2 - 1} + ln \Biggl| \ \frac{x}{|a|} + \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{\left(\frac{x}{|a|} \right)^2 - 1} \ \Biggr| \Biggr] - a^2 \ ln \Biggl| \ \frac{x}{|a|} + \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{\left(\frac{x}{|a|} \right)^2 - 1} \ \Biggr| $$

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x^2}{a^2|x|} \sqrt{x^2 - a^2} + ln \Biggl| \ \frac{x}{|a|} + \frac{x}{|ax|} \sqrt{x^2 - a^2} \ \Biggr| \Biggr] - a^2 \ ln \Biggl| \ \frac{x}{|a|} + \frac{x}{|ax|} \sqrt{x^2 - a^2} \ \Biggr| $$

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x^2}{a^2|x|} \sqrt{x^2 - a^2} + ln \Biggl| \frac{1}{|a|} \left( x + \frac{x \sqrt{x^2 - a^2}}{|x|} \right) \ \Biggr| \Biggr] - a^2 \ ln \Biggl| \frac{1}{|a|} \left( x + \frac{x \sqrt{x^2 - a^2}}{|x|} \right) \ \Biggr| $$

We can now factorize by the big factor containing \(ln|X|\):

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{1}{2} \frac{x^2}{|x|}\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| \frac{1}{|a|} \left( x + \frac{x \sqrt{x^2 - a^2}}{|x|} \right) \ \Biggr| $$

$$ a^2 \int^x sec(u) \ tan^2(u) \ du = \frac{1}{2}\frac{x^2}{|x|}\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| x + \frac{x \sqrt{x^2 - a^2}}{|x|} \ \Biggr| + \frac{ ln|a| a^2}{2} $$




The constant \( + \frac{ ln|a| a^2}{2} \) being absorbed by the main integration constant, we finally obtain by taking in consideration the sign generator left aside:

$$ \int^x K_4(t) \ dt = \frac{tan(u)}{|tan(u)|} \Biggl( \frac{1}{2}\frac{x^2}{|x|}\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| x + \frac{x \sqrt{x^2 - a^2}}{|x|} \ \Biggr| \Biggr) $$

$$ \left(with \ \frac{tan(u)}{|tan(u)|} = \Biggl \{ \begin{align*} = -1, \ \forall x \in \hspace{0.05em} ] -\infty, -a [ \\ = 1, \ \forall x \in \hspace{0.05em} ]a, +\infty [ \end{align*} \right)$$

We then have to cases to manage now:





1. For the negative part: \(] -\infty, -a [ \)

\(x\) is negative and the sign generator too, so:

$$ \forall x \in \hspace{0.05em} ] -\infty, -a [, \ \int^x K_4(t) \ dt = -\Biggl( -\frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| -|x| - \sqrt{x^2 - a^2} \ \Biggr| \Biggr) $$

$$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} + \frac{a^2}{2} ln \Biggl| -|x| - \sqrt{x^2 - a^2} \ \Biggr| $$

$$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl( \frac{1}{\left| -|x| - \sqrt{x^2 - a^2} \right|} \ \Biggr) $$

$$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl( \frac{-|x| + \sqrt{x^2 - a^2}}{a^2} \ \Biggr) $$

$$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| -|x| + \sqrt{x^2 - a^2} \Biggr| +ln(a^2) $$

$$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| x + \sqrt{x^2 - a^2} \Biggr| +2 \ ln(a) $$

The constant \( +2 \ ln(a) \) will be also absorbed by the main integration constant.




2. For the positive part: \(] a, +\infty [ \)

\(x\) is positive and the sign generator too, so:

$$ \forall x \in \hspace{0.05em} ] a, +\infty [, \ \int^x K_4(t) \ dt =\frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} ln \Biggl| x+ \sqrt{x^2 - a^2} \ \Biggr| $$




3. Conclusion

In any case, \((x < -a) \ or \ (x > a)\), this integral is worth:

$$ \Bigl[ \forall a \in \hspace{0.05em} \mathbb{R}^*, \ \forall x \in \bigl] -\infty, -|a| \bigr[ \hspace{0.05em} \cup \hspace{0.05em} \bigl]|a|, +\infty \bigr[ \Bigr] \lor \Bigl[ (x = 0) \land (x > 0) \Bigr],$$

$$ \int^x \sqrt{x^2 - a^2} \ dt = \frac{ x\sqrt{x^2 - a^2} - a^2 \ ln \left| x+ \sqrt{x^2 - a^2} \ \right| }{2} $$

Integration methods and antiderivatives recap table

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Integration examples

1. Integration by parts examples

1. Example 1

$$ \alpha = \int_{0}^{\frac{\pi}{2}} t .sin(t) \hspace{0.2em}dt $$

We choose to take \(f\) and \(g'\) such as:

$$ \Biggl \{ \begin{align*} f(t) = t \\ g'(t) = dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = \frac{dt}{t} \\ g(t) = t.dt \end{align*} $$

$$ \alpha = \Bigl[-t.cos(t) \Bigr]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} - cos(t) \hspace{0.2em}dt $$

$$ \alpha = \Bigl[-t.cos(t) + sin(t) \Bigr]_{0}^{\frac{\pi}{2}} $$

$$ \alpha = - \frac{\pi}{2} \times 0 + 1 - (0\times 1 + 0 ) $$

$$ \alpha =1 $$



2. Example 2

$$ \beta = \int_{1}^{e} ln(t) \hspace{0.2em}dt $$

We choose to take \(f\) and \(g'\) such as:

$$ \Biggl \{ \begin{align*} f(t) = ln(t) \\ g'(t) = dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = \frac{dt}{t} \\ g(t) = t \end{align*} $$

$$ \beta = \Bigl[t.ln(t) \Bigr]_{1}^{e} - \int_{1}^{e} \hspace{0.2em}dt $$

$$ \beta = \Bigl[t.ln(t) \Bigr]_{1}^{e}- \Bigl[t \Bigr]_{1}^{e} $$

$$ \beta = e . ln(e) - ln(1) - (e -1) $$

$$ \beta =1 $$



3. Example 3: integrate a function of type \(f(t) e^t\)

Notably while solving ordinary differential equations, we may determine antiderivatives of type:

$$ \int^{x} f(t) \hspace{0.1em} e^t \hspace{0.2em}dt $$

For instance, with a polynomial function \(f(t)\), we integrate many times in a row until the degree decreases to reach \(0\).

We choose to integrate the exponential function so that the polynomial function is the one which is derivated and decrease in degree as integrations progresses.

Let us integrate this polynomial-exponential function:

$$ \gamma = \int^{x} (4t^3 - t +4 ) \hspace{0.1em} e^{2t} \hspace{0.2em}dt $$

$$ \Biggl \{ \begin{align*} f(t) = 4t^3 - t +4 \\ g'(t) = e^{2t} \hspace{0.2em}dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = (12t^2 -1) \hspace{0.2em}dt \\ g(t) = \frac{e^{2t}}{2} \end{align*} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - \int^{x} ( 12t^2 -1 ) \hspace{0.1em} \frac{e^{2t}}{2} \hspace{0.2em}dt $$

We take this oppotunity to take the constant out of it.

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6\int^{x} t^2 \hspace{0.1em} e^{2t} \hspace{0.2em}dt + \int^{x} \frac{e^{2t}}{2} \hspace{0.2em}dt $$

$$ \Biggl \{ \begin{align*} f(t) = t^2 \\ g'(t) = e^{2t} \hspace{0.2em}dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = 2t \hspace{0.2em}dt \\ g(t) = \frac{e^{2t}}{2} \end{align*} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6 \Biggl( \Biggl[ t^2 \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - \int^{x} t \hspace{0.1em} e^{2t} \hspace{0.2em}dt \Biggr) + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t} \Biggr]^{x} + 6 \int^{x} t \hspace{0.1em} e^{2t} \hspace{0.2em}dt + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \Biggl \{ \begin{align*} f(t) = t \\ g'(t) = e^{2t} \hspace{0.2em}dt \end{align*} $$

$$ \Biggl \{ \begin{align*} f'(t) = dt \\ g(t) = \frac{e^{2t}}{2} \end{align*} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t}\Biggr]^{x} + 6\Biggl[t \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6 \int^{x} \frac{e^{2t}}{2} \hspace{0.2em}dt + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t}\Biggr]^{x} + 3\Biggl[t \hspace{0.1em} e^{2t} \Biggr]^{x} - 3\Biggl[ \frac{e^{2t}}{2} \Biggr]^{x} + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$

$$ \gamma = \frac{1}{2} \left (4x^3 - x +4 \right) \hspace{0.1em} e^{2x} - 3x^2 e^{2x} + 3x e^{2x} - \frac{5}{4}e^{2x} $$

We finally factorize it all.

$$ \gamma = e^{2x} \left (2x^3 - \frac{1}{2}x +2 - 3x^2 + 3x - \frac{5}{4} \right) \hspace{0.1em} $$

$$ \gamma = e^{2x} \left (2x^3 -3x^2 + \frac{5}{2} x + \frac{3}{4} \right) \hspace{0.1em} $$

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2. Integration by substitution examples

1. Example 1

$$ A = \int_{0}^{\frac{\pi}{2}} \frac{sin(t)}{1 + 3 cos^2(t)} \ dt$$

In the case of a trigonometric polynomial function, we can apply Bioche's rules.

In our case, the integrand \( \omega(t) \) is invariant by the variable change \( t = -t \):

$$ \omega(-t) = \frac{sin(-t)}{1 + 3 cos^2(-t)} \ d(-t) $$

$$ \omega(-t) = \frac{-sin(t)}{1 + 3 cos^2(t)} \ -d(t) $$

$$ \omega(-t) = \frac{sin(t)}{1 + 3 cos^2(t)} \ -(t) $$

$$ \omega(-t) = \omega(t) $$

We thus set a new variable down:

$$ u = cos(t)$$

We do have now:

$$ \Biggl \{ \begin{align*} u = cos(t) \\ du = -sin(t) \ dt \end{align*} $$

We replace it in our main expression, without forgetting the bounds:

$$ A = \int_{cos(0)}^{cos(\frac{\pi}{2})} \frac{-du}{1 + 3 u^2} $$

$$ A = -\int_{1}^{0} \frac{1}{1 + 3 u^2} \ du $$

Thanks to this property of the integrals, we know that:

$$ \forall (a,b) \in I^2, \ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$

$$ A = \int_{0}^{1} \frac{1}{1 + 3 u^2} \ du$$

The antiderivative of the arctangent function is known as:

$$ \int^{x} \frac{1}{a^2 + t^2} \ dt = \Biggl[\frac{1}{a} arctan\left( \frac{t}{a}\right) \Biggr]^x $$

Alors, dans notre cas,

$$ A = \frac{1}{3} \int_{0}^{1} \frac{1}{\frac{1}{3} + u^2} \ du$$

Puis,

$$ A = \frac{1}{3} \int_{0}^{1} \frac{1}{ \left(\frac{1}{\sqrt{3}}\right)^2 + \ u^2} \ du = \frac{1}{3} \Bigl[\sqrt{3} \ arctan\left(\sqrt{3} u\right) \Bigr]_0^1 $$

$$ A = \frac{1}{3} \ \Bigl[ \sqrt{3} \ arctan(\sqrt{3}) - \sqrt{3} \ arctan(0) \Bigr] $$

$$ A = \frac{\sqrt{3}}{3} arctan(\sqrt{3}) $$

2. Example 2

$$ B = \int_{0}^{1} \ \frac{t^2}{1 + t^3} \ dt$$

The idea is to set \( u = t^3 \) down to get \( t^2 \ dt\) at the numerator (mutiplied by a constant) after substitution.

$$ \Biggl \{ \begin{align*} u = t^3 \\ du = 3t^2 \ dt \end{align*} $$

We replace it all (bounds do not change because \( 0\) and \( 1 \) do not vary taking their cube):

$$ B = \int_{0}^{1} \ \frac{1}{1 + u^3} \ \frac{du}{3} $$

$$ B = \frac{1}{3} \int_{0}^{1} \ \frac{1}{1 + u} \ du$$

$$ B = \frac{1}{3} \Bigl[ln |1 + u| \Bigr]_{0}^{\ 1} $$

$$ B = \frac{1}{3} ln(2) $$

3. Example 3

$$ C = \int_{1}^{3} \ \frac{e^{2t}}{1 - e^t} \ dt$$

The integrand in only defined for \( \mathbb{R} \ \backslash \{0 \}\).

$$ \Biggl \{ \begin{align*} u = e^t \\ du = e^t \ dt \end{align*} $$

$$ C = \int_{e}^{e^3} \ \frac{u}{1 - u} \ du$$

Let us use a little trick to transform the expression:

$$ C = \int_{e}^{e^3} \ \frac{u + 1 - 1}{1 - u} \ du$$

$$ C = \int_{e}^{e^3} \ \frac{u - 1}{1 - u} + \frac{1}{1 - u} \ du$$

$$ C = \int_{e}^{e^3} \ \frac{-(1-u)}{1 - u} + \frac{1}{1 - u} \ du$$

$$ C = -\int_{e}^{e^3} du + \int_{e}^{e^3} \frac{1}{1 - u} \ du$$

$$ C = - \Bigl[u\Bigr]_{e}^{e^3} + \Bigl[ - ln |1 - u| \Bigr]_{e}^{e^3} $$

$$ C = - e^3 + e - ln |1 - e^3| + ln |1 - e| $$

$$ C = - e^3 + e - ln (e^3-1) + ln(e-1) $$




3. Integration of a rational fraction with a second degree denominator examples\(: {\displaystyle \int^x} \frac{1}{Q_2(t)} \ dt \)

1. Example 1: with a positive discriminant \((\Delta > 0)\)

$$ D = \int^x \frac{dt}{2t^2 -4t + 1} $$

We try to get back to a type of form \(\frac{1}{x^2 + px + q}\).

$$ D = \frac{1}{2} \int^x \frac{dt}{t^2 -2t + \frac{1}{2}} $$

Let's calculate the discriminant \(\Delta\).

$$ \Delta = p^2 - 4q $$

$$ \Delta = (-2)^2 - 4 \times \frac{1}{2} $$

$$ \Delta = 4 - 2 = 2 $$

We then have two solutions \((t_1, t_2)\).

$$ \left \{ \begin{align*} t_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ t_2 = \frac{- p +\sqrt{p^2 - 4q}}{2} \end{align*} \right \} \Longleftrightarrow \left \{ t_1 = 1 - \frac{\sqrt{2}}{2}, \ t_2 = 1 + \frac{\sqrt{2}}{2} \right \}$$




So, our integral can be written in factorized form:

$$ D = \frac{1}{2} \int^x \frac{dt}{\left(t - 1 + \frac{\sqrt{2}}{2} \right) \left(t - 1 - \frac{\sqrt{2}}{2} \right) } $$

Let us break it in simple elements:

Let us set a function \(F(X) \) down:

$$F(X) = \frac{1}{\left(X - 1 + \frac{\sqrt{2}}{2} \right) \left(X - 1 - \frac{\sqrt{2}}{2} \right) } \qquad (F(X))$$

We are looking for two real numbers \( a \) and \(b\) such as:

$$F(X) = \frac{a}{\left(X - 1 + \frac{\sqrt{2}}{2} \right)} + \frac{b}{\left(X - 1 - \frac{\sqrt{2}}{2} \right)}$$




Performing \( \left(X = 2 - \frac{\sqrt{2}}{2}\right)\), we determine \( a \):

$$ \underset{\left(X = 2 - \frac{\sqrt{2}}{2}\right)}{F(X)} \times \left(X - 1 + \frac{\sqrt{2}}{2} \right) = \frac{1}{\left(X - 1 - \frac{\sqrt{2}}{2} \right)}= a \Longrightarrow \left(a = -\frac{1}{\sqrt{2}} \right) $$

Now, performing \( \left(X = 2 + \frac{\sqrt{2}}{2}\right)\), we determine \( b \):

$$ \underset{\left(X = 2 + \frac{\sqrt{2}}{2}\right)}{F(X)} \times \left(X - 1 - \frac{\sqrt{2}}{2} \right) = \frac{1}{\left(X - 1 + \frac{\sqrt{2}}{2} \right)}= b \Longrightarrow \left(b = \frac{1}{\sqrt{2}} \right) $$




In the end, this integral can be written in decomposed form:

$$ D = \frac{1}{2} \int^x -\frac{1}{\sqrt{2}} \frac{1}{\left(t - 1 + \frac{\sqrt{2}}{2} \right) } \ dt + \frac{1}{2} \int^x \frac{1}{\sqrt{2}} \frac{1}{\left(t - 1 - \frac{\sqrt{2}}{2} \right) } \ dt $$




We can now easily integrate it:

$$ D = -\frac{1}{2\sqrt{2}} \Biggl[ ln \left| t - 1 + \frac{\sqrt{2}}{2} \right| \Biggr]^x + \frac{1}{2\sqrt{2}} \Biggl[ ln\left| t - 1 - \frac{\sqrt{2}}{2} \right| \Biggr]^x$$

$$ D = \frac{1}{2\sqrt{2}} \times \left( ln\left| x - 1 - \frac{\sqrt{2}}{2} \right| - ln\left| x - 1 + \frac{\sqrt{2}}{2} \right| \right) $$

$$ D = \frac{1}{2\sqrt{2}} \times ln\left| \frac{x - 1 - \frac{\sqrt{2}}{2}}{x - 1 + \frac{\sqrt{2}}{2}} \right| $$



2. Exemple 2: with a nul discriminant \((\Delta = 0)\)

$$ E = \int^x \frac{dt}{t^2 + 2t + 1} $$

Let us calculate the discriminant \(\Delta\).

$$ \Delta = p^2 - 4q $$

$$ \Delta = 2^2 - 4 \times 1 $$

$$ \Delta = 4 - 4 = 0 $$

We then have a double solution \( t_0 \).

$$t_0 = \frac{- 2}{2} = -1 $$




Now, our integral can be written in factorized form:

$$ E = \int^x \frac{dt}{(t+1)^2} $$

At this stage, we can easily integrate and:

$$ E = \Bigg[ -\frac{1}{t+1} \Biggr]^x $$

$$ E = -\frac{1}{x+1} $$



3. Example 3: with a negative discriminant \((\Delta < 0)\)

$$ F = \int^x \frac{dt}{t^2 + t + 3} $$

Let us calculate the discriminant \(\Delta\).

$$ \Delta = p^2 - 4q $$

$$ \Delta = 1^2 - 4 \times 3 $$

$$ \Delta = 1 - 12 = -11 $$

In this specific case, we will use the canonic form of the polynomial:

$$ F = \int^x \frac{dt}{\left(t + \frac{1}{2} \right)^2 + \left(3 - \frac{1}{4} \right)} $$

$$ F = \int^x \frac{dt}{\left(t + \frac{1}{2} \right)^2 + \frac{11}{4} } $$

$$ F = \int^x \frac{dt}{\left(t + \frac{1}{2} \right)^2 + \sqrt{\frac{11}{4}}^2 } $$

Now, the integral is a standard antiderivative:

$$ F = \frac{1}{\sqrt{\frac{11}{4}}} \times arctan\left( \frac{t + \frac{1}{2}}{\sqrt{\frac{11}{4}}} \right) $$

$$ F = \frac{\sqrt{4}}{\sqrt{11}} \times arctan\left( \frac{\sqrt{4}}{\sqrt{11}} \left(t + \frac{1}{2} \right) \right) $$

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4. Integration of a rational fraction with first degree numerator and second degree denominator example\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)

$$ G = \int^x \frac{3x-5}{-2t^2 + 5t + 6} \ dt $$

As well as before, we try to obtain a form of type \((x^2 + px + q)\) at the denominator.

$$ G = -\frac{1}{2} \int^x \frac{3x-5}{t^2 -\frac{5}{2}t -3} \ dt $$

We use the method seen above in the demonstration.

$$ \frac{Ax + B}{x^2 + px + q} = \frac{\frac{A}{2} (2x) + \frac{Ap}{2} - \frac{Ap}{2} + B}{x^2 + px + q} $$

$$ \frac{Ax + B}{x^2 + px + q} = \frac{\frac{A}{2} \left(2x + \frac{Ap}{2} \right) - \frac{Ap}{2} + B}{x^2 + px + q} $$

That is to say, split the big quotient in two distincts quotients:

$$ \frac{Ax + B}{x^2 + px + q} = \frac{A}{2}\frac{2x + p}{x^2 + px + q} + \frac{B - \frac{Ap}{2}}{x^2 + px + q} $$

And use the following standard antiderivative:

$$ \int^x \frac{u'}{u} \ du = ln |u|$$

$$ G = -\frac{1}{2} \int^x \frac{\frac{3}{2} \times (2x) + \frac{3 \times \left(-\frac{5}{2}\right)}{2} - \frac{3 \times \left(-\frac{5}{2}\right)}{2} -5}{t^2 -\frac{5}{2}t -3} \ dt $$

$$ G = -\frac{1}{2} \left( \int^x \frac{ \frac{3}{2} \left(2x -\frac{5}{2}\right) + \frac{15}{4} -5 }{t^2 -\frac{5}{2}t -3} \ dt \right) $$

We can now split it in two differents quotients.

$$ G = -\frac{1}{2} \left( \frac{3}{2} \int^x \frac{ \left(2x -\frac{5}{2}\right)}{t^2 -\frac{5}{2}t -3} \ dt + \int^x \frac{ \frac{15}{4} -5 }{t^2 -\frac{5}{2}t -3} \ dt \right) $$

$$ G = -\frac{3}{4} \int^x \frac{ 2x -\frac{5}{2} }{t^2 -\frac{5}{2}t -3} \ dt \ - \ \frac{1}{2} \int^x \frac{\frac{15}{4} - 5}{t^2 -\frac{5}{2}t -3} \ dt $$

$$ G = -\frac{3}{4} \Biggl[ ln \left| t^2 -\frac{5}{2}t -3 \right| \Biggr]^x \ + \ \frac{5}{8} \int^x \frac{1}{t^2 -\frac{5}{2}t -3} \ dt $$

$$ G = -\frac{3}{4} ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} G_2 $$




We must now calculate the integral \(G_2\) using the previous methods. We first determine the discriminant \(\Delta\).

$$\Delta = p^2 - 4q$$

$$\Delta = \left(-\frac{5}{2}\right)^2 - 4 \times (-3)$$

$$\Delta = \frac{25}{4} + 12$$

$$\Delta = \frac{25}{4} + \frac{48}{4}$$

$$\Delta = \frac{73}{4} $$

We then have two solutions \((t_1, t_2)\).

$$ \left \{ \begin{align*} t_1 = \frac{\frac{5}{2} - \sqrt{\frac{73}{4}}}{2}\\ t_2 = \frac{\frac{5}{2} + \sqrt{\frac{73}{4}}}{2}\end{align*} \right \} \Longleftrightarrow \left \{ t_1 = \frac{5 - \sqrt{73}}{4}, \ t_2 = \frac{5 +\sqrt{73}}{4} \right \}$$

So, our integral \(G_2\) can be written in a factorized form:

$$ G_2 = \frac{1}{2} \int^x \frac{dt}{\left(t - \left( \frac{5 - \sqrt{73}}{4} \right) \right) \left(t - \left( \frac{5 + \sqrt{73}}{4} \right) \right) } $$




Let us decompose it in simple elements:

Let us set the function \(F(X) \) down:

$$F(X) = \frac{1}{\left(X - \left( \frac{5 - \sqrt{73}}{4} \right) \right) \left(X - \left( \frac{5 + \sqrt{73}}{4} \right) \right) } \qquad (F(X))$$

We are looking for two real numbers \( \alpha \) and \(\beta\)such as:

$$F(X) = \frac{\alpha}{\left(X - \left( \frac{5 - \sqrt{73}}{4} \right) \right)} + \frac{\beta}{ \left(X - \left( \frac{5 + \sqrt{73}}{4} \right) \right)}$$




Performing \( \left(X = \frac{5 - \sqrt{73}}{4} \right) \), we determine \( \alpha \):

$$ \underset{\left(X = \frac{5 - \sqrt{73}}{4} \right)}{F(X)} \times \left(X - \left( \frac{5 - \sqrt{73}}{4} \right) \right) = \frac{1}{\left(X - \left( \frac{5 + \sqrt{73}}{4} \right) \right)}= \alpha \Longrightarrow \left(\alpha = -\frac{2}{\sqrt{73}} \right) $$

Now performing \( \left(X = \frac{5 + \sqrt{73}}{4} \right) \), we determine \( \beta \):

$$ \underset{\left(X = \frac{5 + \sqrt{73}}{4} \right)}{F(X)} \times \left(X - \left( \frac{5 + \sqrt{73}}{4} \right) \right) = \frac{1}{\left(X - \left( \frac{5 - \sqrt{73}}{4} \right) \right)}= \beta \Longrightarrow \left(\beta = \frac{2}{\sqrt{73}} \right) $$




In the end, this integral can be written in a decomposed form:

$$ G_2 = -\frac{2}{\sqrt{73}} \int^x \frac{dt}{\left(t - \left( \frac{5 - \sqrt{73}}{4} \right) \right)} + \frac{2}{\sqrt{73}}\int^x \frac{dt}{ \left(t - \left( \frac{5 + \sqrt{73}}{4} \right) \right)} $$

$$ G_2 = \frac{2}{\sqrt{73}} \left( \Biggl[ ln\left| t - \frac{5 + \sqrt{73}}{4} \right| - ln\left| t - \frac{5 - \sqrt{73}}{4} \right| \Biggr]^x \right) $$

$$ G_2 = \frac{2}{\sqrt{73}} \ ln\left| \frac{x - \frac{5 + \sqrt{73}}{4}}{x - \frac{5 - \sqrt{73}}{4}} \right| $$

$$ G_2 = \frac{2}{\sqrt{73}} \ ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$




Finally, adding the previous result we obtain the final integral:

$$ G = -\frac{3}{4} ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} G_2 $$

$$ G = -\frac{3}{4} ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} \times \frac{2}{\sqrt{73}} \ ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$

$$ G = -\frac{3}{4} \ ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{4\sqrt{73}} \ ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$