Интеграција рационалних функција 2

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Пр.2)   Решити   $\int {\frac{{x - 3}}{{{x^3} - x}}} dx$.

Пр.2)  $\int {\frac{{x - 3}}{{{x^3} - x}}} dx=$

$\frac{{x - 3}}{{{x^3} - x}} = \frac{{x - 3}}{{x\left( {{x^2} - 1} \right)}} = \frac{{x - 3}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{\left( {x - 1} \right)}} + \frac{C}{{\left( {x + 1} \right)}} = $

$ = \frac{{A\left( {x + 1} \right)\left( {x - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}}$

$x - 3 = A\left( {x + 1} \right)\left( {x - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right) = $

$ = A\left( {{x^2} - 1} \right) + B\left( {{x^2} + x} \right) + C\left( {{x^2} - x} \right) = $

$ = A{x^2} - A + B{x^2} + Bx + C{x^2} - Cx = \left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A$

$0 \cdot {x^2} + x - 3 = \left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A$

\[\left\{ \begin{gathered}
A + B + C = 0 \hfill \\
B - C = 1 \hfill \\
- A = - 3 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
B + C = - 3 \hfill \\
B - C = 1 \hfill \\
A = 3 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
2B = - 2 \hfill \\
B - C = 1 \hfill \\
A = 3 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
B = - 1 \hfill \\
C = - 2 \hfill \\
A = 3 \hfill \\
\end{gathered} \right.\]

$\frac{{x - 3}}{{{x^3} - x}} = \frac{3}{x} + \frac{{ - 1}}{{\left( {x - 1} \right)}} + \frac{{ - 2}}{{\left( {x + 1} \right)}}$

$\int {\frac{{x - 3}}{{{x^3} - x}}dx}  = \int {\frac{3}{x}dx}  + \int {\frac{{ - 1}}{{\left( {x - 1} \right)}}dx}  + \int {\frac{{ - 2}}{{\left( {x + 1} \right)}}dx}  = $

$ = 3\int {\frac{1}{x}dx}  - \int {\frac{1}{{\left( {x - 1} \right)}}dx}  - 2\int {\frac{1}{{\left( {x + 1} \right)}}dx}  = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
x - 1 = t \hfill \\
dx = dt \hfill \\
\end{gathered} &{}&\begin{gathered}
x + 1 = s \hfill \\
dx = ds \hfill \\
\end{gathered}
\end{array}\]

$ = 3\int {\frac{1}{x}dx}  - \int {\frac{{dt}}{t}}  - 2\int {\frac{{dt}}{t}}  = 3\ln \left| x \right| - \ln \left| t \right| - 2\ln \left| s \right| + C = $

$ = 3\ln \left| x \right| - \ln \left| {x - 1} \right| - 2\ln \left| {x + 1} \right| + C = \ln {\left| x \right|^3} - \ln \left| {x - 1} \right| - \ln {\left| {x + 1} \right|^2} + C = $

$ = \ln \left| {\frac{{{x^3}}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}} \right| + C$