Integration of Rational Functions
Solved Problems
Example 9.
Find the integral \[\int {\frac{{x + 1}}{{{{\left( {x - 2} \right)}^3}}} dx}.\]
Solution.
We can use the partial fraction decomposition to convert the rational function into an integrable form. The result will be the same if we directly simplify the integral to obtain:
\[\int {\frac{{x + 1}}{{{{\left( {x - 2} \right)}^3}}}dx} = \int {\frac{{x - 2 + 3}}{{{{\left( {x - 2} \right)}^3}}}dx} = \int {\frac{{x - 2}}{{{{\left( {x - 2} \right)}^3}}}dx} + \int {\frac{3}{{{{\left( {x - 2} \right)}^3}}}dx} = \int {\frac{{dx}}{{{{\left( {x - 2} \right)}^2}}}} + 3\int {\frac{{dx}}{{{{\left( {x - 2} \right)}^3}}}} = - \frac{1}{{x - 2}} + 3 \cdot \frac{1}{{\left( { - 2} \right){{\left( {x - 2} \right)}^2}}} + C = - \frac{1}{{x - 2}} - \frac{3}{{2{{\left( {x - 2} \right)}^2}}} + C = \frac{{ - 2\left( {x - 2} \right) - 3}}{{2{{\left( {x - 2} \right)}^2}}} + C = \frac{{ - 2x + 4 - 3}}{{2{{\left( {x - 2} \right)}^2}}} + C = \frac{{1 - 2x}}{{2{{\left( {x - 2} \right)}^2}}} + C.\]
Example 10.
Evaluate the integral \[\int {\frac{{xdx}}{{{x^2} + 2x + 2}}}.\]
Solution.
Given that the derivative of the denominator is
\[\left( {{x^2} + 2x + 2} \right)^\prime = 2x + 2,\]
we rewrite the integral as follows
\[\int {\frac{{xdx}}{{{x^2} + 2x + 2}}} = \frac{1}{2}\int {\frac{{2xdx}}{{{x^2} + 2x + 2}}} = \frac{1}{2}\int {\frac{{2x + 2 - 2}}{{{x^2} + 2x + 2}}dx} = \frac{1}{2}\int {\frac{{2x + 2}}{{{x^2} + 2x + 2}}dx} - \frac{1}{2}\int {\frac{2}{{{x^2} + 2x + 2}}dx} = \frac{1}{2}\int {\frac{{2x + 2}}{{{x^2} + 2x + 2}}dx} - \int {\frac{{dx}}{{{x^2} + 2x + 2}}} = {I_1} - {I_2}.\]
We evaluate the first integral \({I_1}\) by substitution:
\[u = {x^2} + 2x + 2,\;\;du = \left( {2x + 2} \right)dx.\]
Hence
\[{I_1} = \frac{1}{2}\int {\frac{{2x + 2}}{{{x^2} + 2x + 2}}dx} = \frac{1}{2}\int {\frac{{du}}{u}} = \frac{1}{2}\ln \left| u \right| = \frac{1}{2}\ln \left| {{x^2} + 2x + 2} \right| = \frac{1}{2}\ln \left( {{x^2} + 2x + 2} \right).\]
To find the second integral \({I_2}\) we complete the square in the denominator:
\[{I_2} = \int {\frac{{dx}}{{{x^2} + 2x + 2}}} = \int {\frac{{dx}}{{\left( {{x^2} + 2x + 1} \right) + 1}}} = \int {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + 1}}} = \arctan \left( {x + 1} \right).\]
The final answer is
\[I = \frac{1}{2}\ln \left( {{x^2} + 2x + 2} \right) - \arctan \left( {x + 1} \right) + C.\]
Example 11.
Evaluate the integral \[\int {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}}.\]
Solution.
We decompose the integrand into partial functions:
\[\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{x - 3}}.\]
Equate coefficients:
\[A\left( {x - 2} \right)\left( {x - 3} \right) + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right)\left( {x - 2} \right) = {x^2},\]
\[A{x^2} - 2Ax - 3Ax + 6A + B{x^2} - Bx - 3Bx + 3B + C{x^2} - Cx - 2Cx + 2C = {x^2},\]
\[\left( {A + B + C} \right){x^2} - \left( {5A + 4B + 3C} \right)x + 6A + 3B + 2C = {x^2}.\]
Hence,
\[
\left\{ \begin{array}{l}
A + B + C = 1\\
5A + 4B + 3C = 0\\
6A + 3B + 2C = 0
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
A = \frac{1}{2}\\
B = - 4\\
C = \frac{9}{2}
\end{array} \right..\]
Then
\[\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \frac{{\frac{1}{2}}}{{x - 1}} - \frac{4}{{x - 2}} + \frac{{\frac{9}{2}}}{{x - 3}}.\]
The integral is given by
\[\int {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}} = \frac{1}{2}\int {\frac{{dx}}{{x - 1}}} - 4\int {\frac{{dx}}{{x - 2}}} + \frac{9}{2}\int {\frac{{dx}}{{x - 3}}} = \frac{1}{2}\ln \left| {x - 1} \right| - 4\ln \left| {x - 2} \right| + \frac{9}{2}\ln \left| {x - 3} \right| + C.\]
Example 12.
Find the integral \[\int {\frac{{xdx}}{{{{\left( {x + 2} \right)}^3}}}}.\]
Solution.
The linear expression in the denominator has multiplicity \(3,\) so the partial fraction decomposition of the rational function is written as
\[\frac{x}{{{{\left( {x + 2} \right)}^3}}} = \frac{A}{{{{\left( {x + 2} \right)}^3}}} + \frac{B}{{{{\left( {x + 2} \right)}^2}}} + \frac{C}{{x + 2}}.\]
Simplify and equate the coefficients of like powers of \(x:\)
\[x = A + B\left( {x + 2} \right) + C{\left( {x + 2} \right)^2},\]
\[\color{red}x = \color{blue}A + \color{red}{Bx} + \color{blue}{2B} + \color{darkgreen}{C{x^2}} + \color{red}{4Cx} + \color{blue}{4C},\]
\[\color{red}x = \color{darkgreen}{C{x^2}} + \left( \color{red}{B + 4C} \right)\color{red}x + \left( \color{blue}{A + 2B + 4C} \right).\]
We get the system of equations:
\[\left\{ \begin{array}{l}
C = 0\\
B + 4C = 1\\
A + 2B + 4C = 0
\end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l}
C = 0\\
B = 1\\
A = - 2
\end{array} \right..\]
Then
\[\frac{x}{{{{\left( {x + 2} \right)}^3}}} = \frac{{\left( { - 2} \right)}}{{{{\left( {x + 2} \right)}^3}}} + \frac{1}{{{{\left( {x + 2} \right)}^2}}},\]
and we can easily compute the integral:
\[\int {\frac{{xdx}}{{{{\left( {x + 2} \right)}^3}}}} = \int {\frac{{\left( { - 2} \right)dx}}{{{{\left( {x + 2} \right)}^3}}}} + \int {\frac{{dx}}{{{{\left( {x + 2} \right)}^2}}}} = \left( { - 2} \right) \cdot \frac{1}{{\left( { - 2} \right){{\left( {x + 2} \right)}^2}}} - \frac{1}{{x + 2}} + C = \frac{1}{{{{\left( {x + 2} \right)}^2}}} - \frac{1}{{x + 2}} + C = \frac{{1 - x - 2}}{{{{\left( {x + 2} \right)}^2}}} + C = - \frac{{x + 1}}{{{{\left( {x + 2} \right)}^2}}} + C.\]
Example 13.
Evaluate \[\int {\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}}.\]
Solution.
Decompose the integrand into the sum of two fractions:
\[\frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 1}}.\]
Equate coefficients:
\[A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) = 1,\;\;\Rightarrow A{x^2} + A + B{x^2} + Cx + Bx + C = 1,\;\; \Rightarrow \left( {A + B} \right){x^2} + \left( {B + C} \right)x + A + C = 1.\]
Hence
\[
\left\{ \begin{array}{l}
A + B = 0\\
B + C = 0\\
A + C = 1
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
A = \frac{1}{2}\\
B = - \frac{1}{2}\\
C = \frac{1}{2}
\end{array} \right..\]
The integrand can be written as
\[\frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{\frac{1}{2}}}{{x + 1}} + \frac{{ - \frac{1}{2}x + \frac{1}{2}}}{{{x^2} + 1}} = \frac{1}{{2\left( {x + 1} \right)}} - \frac{1}{2} \cdot \frac{x}{{{x^2} + 1}} + \frac{1}{2} \cdot \frac{1}{{{x^2} + 1}}.\]
The initial integral becomes
\[\int {\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}} = \frac{1}{2}\int {\frac{{dx}}{{x + 1}}} - \frac{1}{2}\int {\frac{{xdx}}{{{x^2} + 1}}} + \frac{1}{2}\int {\frac{{dx}}{{{x^2} + 1}}} = \frac{1}{2}\ln \left| {x + 1} \right| - \frac{1}{4}\int {\frac{{d\left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} + \frac{1}{2}\arctan x = \frac{1}{2}\ln \left| {x + 1} \right| - \frac{1}{4}\ln \left( {{x^2} + 1} \right) + \frac{1}{2}\arctan x + C.\]
Example 14.
Evaluate the integral \[\int {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}}}.\]
Solution.
Using the partial fraction decomposition, we get
\[\frac{1}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{{{x^2}}} + \frac{B}{x} + \frac{C}{{x + 1}}.\]
Determine the unknown coefficients:
\[1 = A\left( {x + 1} \right) + Bx\left( {x + 1} \right) + C{x^2},\]
\[\color{blue}1 = \color{red}{Ax} + \color{blue}{A} + \color{darkgreen}{B{x^2}} + \color{red}{Bx} + \color{darkgreen}{C{x^2}},\]
\[\color{blue}1 = \left( \color{darkgreen}{B + C} \right)\color{darkgreen}{x^2} + \left( \color{red}{A + B} \right)\color{red}x + \color{blue}{A}.\]
Hence
\[\left\{ \begin{array}{l}
B + C = 0\\
A + B = 0\\
A = 1
\end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l}
A = 1\\
B = - 1\\
C = 1
\end{array} \right..\]
The rational function is written as the sum of following partial fractions:
\[\frac{1}{{{x^2}\left( {x + 1} \right)}} = \frac{1}{{{x^2}}} - \frac{1}{x} + \frac{1}{{x + 1}}.\]
So the integral is given by
\[\int {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}}} = \int {\left( {\frac{1}{{{x^2}}} - \frac{1}{x} + \frac{1}{{x + 1}}} \right)dx} = \int {\frac{{dx}}{{{x^2}}}} - \int {\frac{{dx}}{x}} + \int {\frac{{dx}}{{x + 1}}} = - \frac{1}{x} - \ln \left| x \right| + \ln \left| {x + 1} \right| + C = \ln \left| {\frac{{x + 1}}{x}} \right| - \frac{1}{x} + C.\]
Example 15.
Find the integral \[\int {\frac{{dx}}{{{x^3} + 1}}}.\]
Solution.
We can factor the denominator in the integrand:
\[{x^3} + 1 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right).\]
Decompose the integrand into partial functions:
\[\frac{1}{{{x^3} + 1}} = \frac{1}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} - x + 1}}.\]
Equate coefficients
\[A\left( {{x^2} - x + 1} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) = 1,\;\; \Rightarrow A{x^2} - Ax + A + B{x^2} + Cx + Bx + C = 1,\;\; \Rightarrow \left( {A + B} \right){x^2} + \left( { - A + B + C} \right)x + A + C = 1.\]
Hence,
\[
\left\{ \begin{array}{l}
A + B = 0\\
- A + B + C = 0\\
A + C = 1
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
A = \frac{1}{3}\\
B = - \frac{1}{3}\\
C = \frac{2}{3}
\end{array} \right..\]
Then
\[\frac{1}{{{x^3} + 1}} = \frac{{\frac{1}{3}}}{{x + 1}} + \frac{{ - \frac{1}{3}x + \frac{2}{3}}}{{{x^2} - x + 1}} = \frac{1}{{3\left( {x + 1} \right)}} - \frac{1}{3} \cdot \frac{{x - 2}}{{{x^2} - x + 1}}.\]
Now we can calculate the initial integral:
\[\int {\frac{{dx}}{{{x^3} + 1}}} = \frac{1}{3}\int {\frac{{dx}}{{x + 1}}} - \frac{1}{3}\int {\frac{{x - 2}}{{{x^2} - x + 1}}dx} = \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{3}\int {\frac{{x - \frac{1}{2} - \frac{3}{2}}}{{{x^2} - x + 1}}dx} = \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{3}\int {\frac{{x - \frac{1}{2}}}{{{x^2} - x + 1}}dx} + \frac{1}{2}\int {\frac{{dx}}{{{x^2} - x + 1}}} = \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{6}\int {\frac{{\left( {2x - 1} \right)dx}}{{{x^2} - x + 1}}} + \frac{1}{2}\int {\frac{{dx}}{{{{\left( {x - \frac{1}{2}} \right)}^2} + \frac{3}{4}}}} = \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{6}\int {\frac{{d\left( {{x^2} - x + 1} \right)}}{{{x^2} - x + 1}}} + \frac{1}{2}\int {\frac{{d\left( {x - \frac{1}{2}} \right)}}{{{{\left( {x - \frac{1}{2}} \right)}^2} } + {{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = \frac{1}{3}\ln \left| {x + 1} \right| - \frac{1}{6}\ln \left( {{x^2} - x + 1} \right) + \frac{1}{{\sqrt 3 }}\arctan \frac{{2x - 1}}{{\sqrt 3 }} + C.\]
Example 16.
Evaluate the integral \[\int {\frac{{dx}}{{{x^4} - 1}}}.\]
Solution.
We can factor the denominator in the integrand:
\[{x^4} - 1 = \left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) = \left( {x - 1} \right)\cdot\left( {x + 1} \right)\cdot \left( {{x^2} + 1} \right).\]
Decompose the integrand into partial functions:
\[\frac{1}{{{x^4} - 1}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{{Cx + D}}{{{x^2} + 1}}.\]
Equate coefficients:
\[
A\left( {{x^2} + 1} \right)\left( {x + 1} \right) + B\left( {{x^2} + 1} \right)\left( {x - 1} \right) + \left( {Cx + D} \right)\left( {{x^2} - 1} \right) = 1,\]
\[A{x^3} + Ax + A{x^2} + A + B{x^3} - B{x^2} + Bx - B + C{x^3} + D{x^2} - Cx - D = 1,\]
\[\left( {A + B + C} \right){x^3} + \left( {A - B + D} \right){x^2} + \left( {A + B - C} \right)x + A - B - D = 1.\]
Hence,
\[
\left\{ \begin{array}{l}
A + B + C = 0\\
A - B + D = 0\\
A + B - C = 0\\
A - B - D = 1
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
A = \frac{1}{4}\\
B = - \frac{1}{4}\\
C = 0\\
D = - \frac{1}{2}
\end{array} \right..\]
Thus, the integrand becomes
\[\frac{1}{{{x^4} - 1}} = \frac{{\frac{1}{4}}}{{x - 1}} - \frac{{\frac{1}{4}}}{{x + 1}} - \frac{{\frac{1}{2}}}{{{x^2} + 1}}.\]
So, the complete answer is
\[\int {\frac{{dx}}{{{x^4} - 1}}} = \frac{1}{4}\int {\frac{{dx}}{{x - 1}}} - \frac{1}{4}\int {\frac{{dx}}{{x + 1}}} - \frac{1}{2}\int {\frac{{dx}}{{{x^2} + 1}}} = \frac{1}{4}\ln \left| {x - 1} \right| - \frac{1}{4}\ln \left| {x + 1} \right| - \frac{1}{2}\arctan x + C = \frac{1}{4}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}\arctan x + C.\]
Example 17.
Calculate the integral \[\int {{\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}}} dx}.\]
Solution.
We decompose the integrand into partial functions, taking into account that the denominator has a third degree root:
\[\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}} = \frac{A}{{{{\left( {x - 1} \right)}^3}}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x - 1}}.\]
Equate coefficients:
\[A + B\left( {x - 1} \right) + C{\left( {x - 1} \right)^2} = 5x,\;\; \Rightarrow A + Bx - B + C{x^2} - 2Cx + C = 5x,\;\; \Rightarrow C{x^2} + \left( {B - 2C} \right)x + A - B + C = 5x.\]
We get the following system of equations:
\[
\left\{ \begin{array}{l}
C = 0\\
B - 2C = 5\\
A - B + C = 0
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
A = 5\\
B = 5\\
C = 0
\end{array} \right..\]
Hence,
\[\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}} = \frac{5}{{{{\left( {x - 1} \right)}^3}}} + \frac{5}{{{{\left( {x - 1} \right)}^2}}}.\]
The initial integral is equal to
\[\int {\frac{{5x}}{{{{\left( {x - 1} \right)}^3}}}dx} = \int {\left( {\frac{5}{{{{\left( {x - 1} \right)}^3}}} + \frac{5}{{{{\left( {x - 1} \right)}^2}}}} \right)dx} = 5\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^3}}}} + 5\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} = 5 \cdot \frac{{{{\left( {x - 1} \right)}^{ - 2}}}}{{ - 2}} - \frac{5}{{x - 1}} + C = - \frac{5}{{2{{\left( {x - 1} \right)}^2}}} - \frac{5}{{x - 1}} + C.\]
Example 18.
Evaluate the integral \[\int {\frac{{dx}}{{{{\left( {{x^2} - 1} \right)}^2}}}}.\]
Solution.
The partial fraction decomposition of the rational function has the form
\[\frac{1}{{{{\left( {{x^2} - 1} \right)}^2}}} = \frac{1}{{{{\left( {x - 1} \right)}^2}{{\left( {x + 1} \right)}^2}}} = \frac{A}{{{{\left( {x - 1} \right)}^2}}} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x + 1} \right)}^2}}} + \frac{D}{{x + 1}}.\]
Calculate the unknown coefficients \(A, B, C, D:\)
\[1 = A{\left( {x + 1} \right)^2} + B\left( {x - 1} \right){\left( {x + 1} \right)^2} + C{\left( {x - 1} \right)^2} + D\left( {x + 1} \right){\left( {x - 1} \right)^2},\]
\[1 = A\left( {{x^2} + 2x + 1} \right) + \left( {Bx - B} \right)\left( {{x^2} + 2x + 1} \right) + C\left( {{x^2} - 2x + 1} \right) + \left( {Dx + D} \right)\left( {{x^2} - 2x + 1} \right),\]
\[\color{blue}1 = \color{darkgreen}{A{x^2}} + \color{red}{2Ax} + \color{blue}A + \color{magenta}{B{x^3}} - \color{darkgreen}{B{x^2}} + \color{darkgreen}{2B{x^2}} - \color{red}{2Bx} + \color{red}{Bx} - \color{blue}B + \color{darkgreen}{C{x^2}} - \color{red}{2Cx} + \color{blue}C + \color{magenta}{D{x^3}} + \color{darkgreen}{D{x^2}} - \color{darkgreen}{2D{x^2}} - \color{red}{2Dx} + \color{red}{Dx} + \color{blue}D,\]
\[\color{blue}1 = \left( \color{magenta}{B + D} \right)\color{magenta}{x^3} + \left( \color{darkgreen}{A + B + C - D} \right)\color{darkgreen}{x^2} + \left( \color{red}{2A - B - 2C - D} \right)\color{red}x + \left( \color{blue}{A - B + C + D} \right).\]
The coefficients can be determined from the system of equations
\[\left\{ \begin{array}{l} B + D = 0\\ A + B + C - D = 0\\ 2A - B - 2C - D = 0\\ A - B + C + D = 1 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} A = \frac{1}{4}\\ B = - \frac{1}{4}\\ C = \frac{1}{4}\\ D = \frac{1}{4} \end{array} \right..\]
Hence, the integrand can be written as the sum of the partial fractions:
\[\frac{1}{{{{\left( {{x^2} - 1} \right)}^2}}} = \frac{1}{{4{{\left( {x - 1} \right)}^2}}} - \frac{1}{{4\left( {x - 1} \right)}} + \frac{1}{{4{{\left( {x + 1} \right)}^2}}} + \frac{1}{{4\left( {x + 1} \right)}}.\]
Integrating these expressions yields:
\[\int {\frac{{dx}}{{{{\left( {{x^2} - 1} \right)}^2}}}} = \int {\frac{{dx}}{{4{{\left( {x - 1} \right)}^2}}}} - \int {\frac{{dx}}{{4\left( {x - 1} \right)}}} + \int {\frac{{dx}}{{4{{\left( {x + 1} \right)}^2}}}} + \int {\frac{{dx}}{{4\left( {x + 1} \right)}}} = \frac{1}{4}\int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2}}}} - \frac{1}{4}\int {\frac{{dx}}{{x - 1}}} + \frac{1}{4}\int {\frac{{dx}}{{{{\left( {x + 1} \right)}^2}}}} + \frac{1}{4}\int {\frac{{dx}}{{x + 1}}} = - \frac{1}{{4\left( {x - 1} \right)}} - \frac{1}{4}\ln \left| {x - 1} \right| - \frac{1}{{4\left( {x + 1} \right)}} + \frac{1}{4}\ln \left| {x + 1} \right| + C = \frac{1}{4}\ln \left| {\frac{{x + 1}}{{x - 1}}} \right| - \frac{1}{4}\left( {\frac{1}{{x - 1}} + \frac{1}{{x + 1}}} \right) + C = \frac{1}{4}\ln \left| {\frac{{x + 1}}{{x - 1}}} \right| - \frac{1}{4} \cdot \frac{{2x}}{{{x^2} - 1}} + C = \frac{1}{4}\ln \left| {\frac{{x + 1}}{{x - 1}}} \right| - \frac{x}{{2{x^2} - 2}} + C.\]
Example 19.
Find the integral \[\int {\frac{{dx}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}}.\]
Solution.
Complete the square in the denominator \({{x^2} + x + 1}:\)
\[\int {\frac{{dx}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}} = \int {\frac{{dx}}{{{{\left( {{x^2} + x + \frac{1}{4} + \frac{3}{4}} \right)}^2}}}} = \int {\frac{{dx}}{{{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}^2}}}} .\]
Now, we can compute the integral using the reduction formula:
\[
\int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} = \frac{t}{{2{m^2}\left( {k - 1} \right){{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}} + \frac{{2k - 3}}{{2{m^2}\left( {k - 1} \right)}} \cdot \int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^{k - 1}}}}}. \]
Then
\[\int {\frac{{dx}}{{{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}^2}}}} = \frac{1}{{2 \cdot \frac{3}{4} \cdot 1 \cdot \left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}} + \frac{{4 - 3}}{{2 \cdot \frac{3}{4} \cdot 1}} \cdot \int {\frac{{dx}}{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}}} = \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{2}{3}\int {\frac{{dx}}{{\left( {{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \right)}}} = \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{2}{3} \cdot \frac{2}{{\sqrt 3 }}\arctan \frac{{x + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} + C = \frac{2}{{3\left( {{x^2} + x + 1} \right)}} + \frac{4}{{3\sqrt 3 }}\arctan \frac{{2x + 1}}{{\sqrt 3 }} + C.\]