Integration methods for rational fractions

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

• \( \alpha) \) 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\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)

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

Integration methods and antiderivatives recap table

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Demonstrations

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,

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

Which now allows us to easily integrate it:

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\(: {\displaystyle \int^x} \frac{1}{Q_2(t)} \ dt \)

• \( \alpha) \) 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:

$$ \left(x + \frac{p}{2}\right)^2 \textcolor{#8E5B5B}{ \ + q - \frac{p^2}{4}} = x^2 + px \textcolor{#8E5B5B}{ + q - \frac{p^2}{4}} + \frac{p^2}{4} $$

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

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\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)

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

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

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.

And finally,

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

Integration methods and antiderivatives recap table

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Integration examples

1. 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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2. 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:

$$ G = -\frac{3}{4} ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} G_2 $$
$$ G = -\frac{3}{4} ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} \times \frac{2}{\sqrt{73}} \ ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$
$$ G = -\frac{3}{4} \ ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{4\sqrt{73}} \ ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$