Index
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\( \alpha) \) Discriminant is positive\(: \Delta > 0 \)
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{x_1, x_2 \bigr \} \Bigr],$$$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{(x_1 - x_2)} \ \ln \left| \frac{x-x_1}{x-x_2} \right|, \hspace{2em}$$$$ \text{with } \left \{ \begin{gather*} x_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ x_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} \end{gather*} \right \} \qquad \bigl(\text{if }\Delta = p^2 - 4q > 0 \bigr) $$
In the specific case where \((p=0, \ q-1)\), we do have as a bonus an explicite definition of the \(\operatorname{Argtanh}\) function:
$$ \forall x \in \hspace{0.04em} ]-1, \hspace{0.1em} 1[, $$$$\operatorname{Argtanh}(x) = \frac{1}{2} \ln \left|\frac{x+1}{x-1} \right|$$ -
\( \beta) \) Discriminant is zero\(: \Delta = 0 \)
$$\forall x \in \hspace{0.04em} \biggl[ \mathbb{R} \hspace{0.04em} \backslash \hspace{0.2em} \Bigl \{ -\frac{p}{2} \Bigr \} \biggr],$$$$ \int^x \frac{1}{t^2 + pt + q} \ dt = - \frac{1}{ x + \frac{p}{2} } \qquad \bigl(\text{if }\Delta = p^2 - 4q = 0 \bigr) $$
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\( \gamma) \) Discriminant is negative\(: \Delta < 0 \)
$$\forall x \in \mathbb{R},$$$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{ \sqrt{ q - \frac{p^2}{4} } } \ \operatorname{Arctan} \left (\frac{x + \frac{p}{2}}{ \sqrt{ q - \frac{p^2}{4} } } \right ) \qquad \bigl(\text{if }\Delta = p^2 - 4q < 0 \bigr) $$
With a numerator of the first degree and a denominator of the second degree\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)
Proofs
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\(: {\displaystyle \int^x} \frac{1}{Q_2(t)} \ dt \)
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\( \alpha) \) Discriminant is positive\(: \Delta > 0 \)
$$ S_1(x) = \frac{1}{x^2 + px + q} \qquad \bigl(\text{if }\Delta = p^2 - 4q > 0 \bigr) $$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:
$$ P_2(X) = (X - X_1)(X - X_2) $$So in our case,
$$ S_1(x) = \frac{1}{(x - x_1) (x - x_2) }$$$$ \text{with } \left \{ \begin{gather*} x_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ x_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} \end{gather*} \right \} $$$$ S_1(x) = \frac{1}{ (x - x_1) (x - x_2) }$$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,
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{x_1, x_2 \bigr \} \Bigr],$$$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{(x_1 - x_2)} \ \ln \left| \frac{x-x_1}{x-x_2} \right|, \hspace{2em}$$$$ \text{with } \left \{ \begin{gather*} x_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ x_2 = \frac{- p + \sqrt{p^2 - 4q}}{2} \end{gather*} \right \} \qquad \bigl(\text{if }\Delta = p^2 - 4q > 0 \bigr) $$
Now, if we study the specific case where\((p = 0, \ q=-1)\), we do have:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.1em}\backslash \hspace{0.2em} \bigl \{ 1, -1 \bigr \} \Bigr] , \ S_{1\alpha}(x) = \frac{1}{x^2 - 1} = \frac{1}{(x-1)(x+1)} $$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 :
$$ \int^x \frac{1}{1 - x^2} \ dt = \frac{1}{2} \bigl( \ln(x+1) - \ln(x-1) \bigr) $$$$ \forall x \in \hspace{0.04em} \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ -1, 1 \bigr \} \Bigr], \ \int^x \frac{1}{1 - x^2} \ dt = \frac{1}{2} \ln \left|\frac{x+1}{x-1} \right| \qquad(2) $$But, we know from the derivatives of trigonometric functions that:
$$ \forall x \in \hspace{0.04em} ]-1, \hspace{0.1em} 1[, \ \operatorname{Argtanh}(x)' = \frac{1}{1 - x^2}$$$$ \forall x \in \hspace{0.04em} ]-1, \hspace{0.1em} 1[, \ \operatorname{Arctan}(x) + C = \int^x \frac{1}{1 - x^2} \ dt \qquad(3) $$Both expressions \((2)\) and \((3)\) having a common member, they are equal up to a constant, respectively in the interval in common.
So,
$$ \forall x \in \hspace{0.04em} ]-1, \hspace{0.1em} 1[, $$$$ \operatorname{Argtanh}(x) + C_1 = \frac{1}{2} \ln \left|\frac{x+1}{x-1} \right| + C_2 $$Let us determine this constant by taking a value of \(x\), for example \(x = 0\).
$$ \operatorname{Argtanh}(0) = 0 $$$$\frac{1}{2} \ln \left|\frac{0+1}{0-1} \right| = \ln(1) - \ln(1) = 0 $$So, we find that:
$$ C_1 = C_2 \Longrightarrow C_1 - C_2 = 0 $$$$ \operatorname{Argtanh}(x) = \frac{1}{2} \ln \left|\frac{x+1}{x-1} \right| $$We then obtain an explicit definition of the \(\operatorname{Argtanh}\) function:
$$ \forall x \in \hspace{0.04em} ]-1, \hspace{0.1em} 1[, $$$$\operatorname{Argtanh}(x) = \frac{1}{2} \ln \left|\frac{x+1}{x-1} \right|$$ -
\( \beta) \) Discriminant is zero\(: \Delta = 0 \)
$$ S_1(x) = \frac{1}{x^2 + px + q} \qquad \bigl(\text{if }\Delta = p^2 - 4q = 0 \bigr) $$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:
$$ P_2(X) = (X - X_0)^2 $$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,
$$\forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \left \{ - \frac{p}{2} \right \} \biggr],$$$$ \int^x \frac{1}{t^2 + pt + q} \ dt = - \frac{1}{ x + \frac{p}{2} } \qquad \bigl(\text{if }\Delta = p^2 - 4q = 0 \bigr) $$ -
\( \gamma) \) Discriminant is negative\(: \Delta < 0 \)
$$ S_1(x) = \frac{1}{x^2 + px + q} \qquad \bigl(\text{if }\Delta = p^2 - 4q < 0 \bigr) $$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:
$$ \left(x + \frac{p}{2}\right)^2 \textcolor{rgb(232 124 124)}{ \ + q - \frac{p^2}{4}} = x^2 + px \textcolor{rgb(232 124 124)}{ + q - \frac{p^2}{4}} + \frac{p^2}{4} $$$$ \left(x + \frac{p}{2}\right)^2 + q - \frac{p^2}{4} = x^2 + px + q $$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:
$$\int^x \frac{u'(t)}{a^2 + u^2} \ dt = \frac{1}{a}\operatorname{Arctan} \left (\frac{u}{a} \right ) $$$$ \text{with } \left \{ \begin{gather*} u(t) =\left(x + \frac{p}{2}\right) \\ u'(t) = 1 \\ a = \sqrt{ q - \frac{p^2}{4} } \end{gather*} \right \} $$So,
$$ \int^x S_1(t) \ dt = \int^x \frac{1}{\left(t + \frac{p}{2}\right)^2 + \Bigl( q - \frac{p^2}{4} \Bigr) } \ dt $$$$ \int^x S_1(t) \ dt = \frac{1}{ \sqrt{ q - \frac{p^2}{4} } } \ \operatorname{Arctan} \left (\frac{x + \frac{p}{2}}{ \sqrt{ q - \frac{p^2}{4} } } \right ) $$As a result we do have,
$$\forall x \in \mathbb{R},$$$$ \int^x \frac{1}{t^2 + pt + q} \ dt = \frac{1}{ \sqrt{ q - \frac{p^2}{4} } } \ \operatorname{Arctan} \left (\frac{x + \frac{p}{2}}{ \sqrt{ q - \frac{p^2}{4} } } \right ) \qquad \bigl(\text{if }\Delta = p^2 - 4q < 0 \bigr) $$
With a numerator of the first degree and a denominator of the second degree\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)
Given that the numerator is almost the derivative of the numerator, let's try to obtain a form like:
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,
Integration methods and antiderivatives recap table
Examples
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Integration of a rational fraction with a second degree denominator examples\(: {\displaystyle \int^x} \frac{1}{Q_2(t)} \ dt \)
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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 \((t_1, t_2)\).
$$ \left \{ \begin{gather*} t_1 = \frac{- p - \sqrt{p^2 - 4q}}{2}\\ t_2 = \frac{- p +\sqrt{p^2 - 4q}}{2} \end{gather*} \right \} \Longleftrightarrow \left \{ t_1 = 1 - \frac{\sqrt{2}}{2}, \ t_2 = 1 + \frac{\sqrt{2}}{2} \right \}$$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 \( \left(X = 1 - \frac{\sqrt{2}}{2}\right)\), we determine \( a \):
$$ \underset{\left(X = 1 - \frac{\sqrt{2}}{2}\right)}{F(X)} \times \left(X - 1 + \frac{\sqrt{2}}{2} \right) = \frac{1}{\left(X - 1 - \frac{\sqrt{2}}{2} \right)}= a \Longrightarrow \left(a = -\frac{1}{\sqrt{2}} \right) $$Now, performing \( \left(X = 1 + \frac{\sqrt{2}}{2}\right)\), we determine \( b \):
$$ \underset{\left(X = 1 + \frac{\sqrt{2}}{2}\right)}{F(X)} \times \left(X - 1 - \frac{\sqrt{2}}{2} \right) = \frac{1}{\left(X - 1 + \frac{\sqrt{2}}{2} \right)}= b \Longrightarrow \left(b = \frac{1}{\sqrt{2}} \right) $$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:
$$ D = -\frac{1}{2\sqrt{2}} \Biggl[ \ln \left| t - 1 + \frac{\sqrt{2}}{2} \right| \Biggr]^x + \frac{1}{2\sqrt{2}} \Biggl[ \ln\left| t - 1 - \frac{\sqrt{2}}{2} \right| \Biggr]^x$$$$ D = \frac{1}{2\sqrt{2}} \times \left( \ln\left| x - 1 - \frac{\sqrt{2}}{2} \right| - \ln\left| x - 1 + \frac{\sqrt{2}}{2} \right| \right) $$$$ D = \frac{1}{2\sqrt{2}} \times \ln\left| \frac{x - 1 - \frac{\sqrt{2}}{2}}{x - 1 + \frac{\sqrt{2}}{2}} \right| $$ -
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:
$$ E = \Bigg[ -\frac{1}{t+1} \Biggr]^x $$$$ E = -\frac{1}{x+1} $$ -
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:
$$ F = \frac{1}{\sqrt{\frac{11}{4}}} \times \operatorname{Arctan}\left( \frac{t + \frac{1}{2}}{\sqrt{\frac{11}{4}}} \right) $$$$ F = \frac{\sqrt{4}}{\sqrt{11}} \times \operatorname{Arctan}\left( \frac{\sqrt{4}}{\sqrt{11}} \left(t + \frac{1}{2} \right) \right) $$
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Example 1: with a positive discriminant \((\Delta > 0)\)
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Integration of a rational fraction with first degree numerator and second degree denominator example\(: {\displaystyle \int^x} \frac{P_1(t)}{Q_2(t)} \ dt \)
$$ G = \int^x \frac{3t-5}{-2t^2 + 5t + 6} \ dt $$As well as before, we try to obtain a form of type \((x^2 + px + q)\) at the denominator.
$$ G = -\frac{1}{2} \int^x \frac{3t-5}{t^2 -\frac{5}{2}t -3} \ dt $$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 \((t_1, t_2)\).
$$ \left \{ \begin{gather*} t_1 = \frac{\frac{5}{2} - \sqrt{\frac{73}{4}}}{2}\\ t_2 = \frac{\frac{5}{2} + \sqrt{\frac{73}{4}}}{2}\end{gather*} \right \} \Longleftrightarrow \left \{ t_1 = \frac{5 - \sqrt{73}}{4}, \ t_2 = \frac{5 +\sqrt{73}}{4} \right \}$$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 :
$$ G = -\frac{3}{4} \ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} G_2 $$$$ G = -\frac{3}{4} \ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{8} \times \frac{2}{\sqrt{73}} \ \ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$$$ G = -\frac{3}{4} \ \ln \left| x^2 -\frac{5}{2}x -3 \right| \ + \ \frac{5}{4\sqrt{73}} \ \ln\left| \frac{4x - (5 + \sqrt{73})}{4x - ( 5 - \sqrt{73})} \right| $$
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