Integration methods for rational fractions with square roots

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

Demonstrations

We seen that with derivatives of trigonometric functions we do have this:

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.

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:

$$ \forall (f,f^{-1}), \enspace (f' \circ f^{-1}) \neq 0, $$

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

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,

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

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

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:

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

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

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