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

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

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

$ = \int {dx + \int {\frac{{{x^2} + 1}}{{{x^3} - {x^2}}}dx = } } $

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

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

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

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

$ = \int {dx}  + \int {\frac{{ - 1}}{x}dx + } \int {\frac{{ - 1}}{{{x^2}}}dx + } \int {\frac{2}{{x - 1}}} dx = $

$ = \int {dx}  - \int {\frac{1}{x}dx - } \int {\frac{1}{{{x^2}}}dx + } 2\int {\frac{1}{{x - 1}}} dx = $

$x - 1 = t$

$dx = dt$

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

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

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