Интеграција рационалних функција 3

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Пр.3)   Решити   $\int {\frac{{2{x^2} - 5x + 1}}{{{x^3} - 2{x^2} + x}}} dx$

Пр.3)   $\int {\frac{{2{x^2} - 5x + 1}}{{{x^3} - 2{x^2} + x}}} =dx$

$\frac{{2{x^2} - 5x + 1}}{{{x^3} - 2{x^2} + x}} = \frac{{2{x^2} - 5x + 1}}{{x\left( {{x^2} - 2x + 1} \right)}} = \frac{{2{x^2} - 5x + 1}}{{x{{\left( {x - 1} \right)}^2}}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x - 1} \right)}^2}}} = $

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

$2{x^2} - 5x + 1 = A{\left( {x - 1} \right)^2} + Bx\left( {x - 1} \right) + Cx =$

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

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

\[\left\{ \begin{gathered}
A + B = 2 \hfill \\
- 2A - B + C = - 5 \hfill \\
A = 1 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
1 + B = 2 \hfill \\
- 2A - B + C = - 5 \hfill \\
A = 1 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
B = 1 \hfill \\
C = - 2 \hfill \\
A = 1 \hfill \\
\end{gathered} \right.\]

$ = \int {\frac{{2{x^2} - 5x + 1}}{{{x^3} - 2{x^2} + x}}} dx = \int {\left( {\frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x - 1} \right)}^2}}}} \right)} dx = \int {\left( {\frac{1}{x} + \frac{1}{{x - 1}} + \frac{{ - 2}}{{{{\left( {x - 1} \right)}^2}}}} \right)} dx = $

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

$x - 1 = t$

$dx = dt$

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

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