Интеграција рационалних функција 1

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Пр.1)   Решити   $\int {\frac{{2x + 7}}{{{x^2} - x - 2}}} $

Пр.1)   $\int {\frac{{2x + 7}}{{{x^2} - x - 2}}} =$

$\frac{{2x + 7}}{{{x^2} - x - 2}} = \frac{{2x + 7}}{{\left( {x + 1} \right)\left( {x - 2} \right)}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{\left( {x - 2} \right)}} = \frac{{A\left( {x - 2} \right) + B\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}$

\[\begin{gathered}
{x^2} - x - 2 = 0 \hfill \\
{x_{1/2}} = \frac{{1 \pm \sqrt {1 + 8} }}{2} = \frac{{1 \pm 3}}{2} \hfill \\
\begin{array}{*{20}{c}}
{{x_1} = - 1}&{}&{{x_2} = 2}
\end{array} \hfill \\
\end{gathered} \]

$\frac{{2x + 7}}{{{x^2} - x - 2}} = \frac{{A\left( {x - 2} \right) + B\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}$

$2x + 7 = A\left( {x - 2} \right) + B\left( {x + 1} \right) = Ax - 2A + Bx + B =$

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

\[\left\{ \begin{gathered}
A + B = 2/ \cdot \left( { - 1} \right) \hfill \\
- 2A + B = 7 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
- 3A = 5 \hfill \\
A + B = 2 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
A = - \frac{5}{3} \hfill \\
- \frac{5}{3} + B = 2 \hfill \\
\end{gathered} \right.\left\{ \begin{gathered}
A = - \frac{5}{3} \hfill \\
B = \frac{{11}}{3} \hfill \\
\end{gathered} \right.\]

$\frac{{2x + 7}}{{{x^2} - x - 2}} = \frac{{ - \frac{5}{3}}}{{\left( {x + 1} \right)}} + \frac{{\frac{{11}}{3}}}{{\left( {x - 2} \right)}}$

$\int {\frac{{2x + 7}}{{{x^2} - x - 2}}dx}  = \int {\frac{{ - \frac{5}{3}}}{{\left( {x + 1} \right)}}dx}  + \int {\frac{{\frac{{11}}{3}}}{{\left( {x - 2} \right)}}dx}  = $

$ =  - \frac{5}{3}\int {\frac{{dx}}{{\left( {x + 1} \right)}}}  + \frac{{11}}{3}\int {\frac{{dx}}{{\left( {x - 2} \right)}}}  = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
x + 1 = t \hfill \\
dx = dt \hfill \\
\end{gathered} &{}&\begin{gathered}
x - 2 = s \hfill \\
dx = ds \hfill \\
\end{gathered}
\end{array}\]

$ =  - \frac{5}{3}\int {\frac{{dt}}{t}}  + \frac{{11}}{3}\int {\frac{{ds}}{s}}  =  - \frac{5}{3}\ln \left| t \right| + \frac{{11}}{3}\ln \left| s \right| + C =  - \frac{5}{3}\ln \left| {x + 1} \right| + \frac{{11}}{3}\ln \left| {x - 2} \right| + C = $

$ = \ln {\left| {x + 1} \right|^{ - \frac{5}{3}}} + \ln {\left| {x - 2} \right|^{\frac{{11}}{3}}} + C = \ln \frac{1}{{{{\sqrt[3]{{x + 1}}}^5}}} + \ln {\sqrt[3]{{x - 2}}^{11}} + C = $

$ = \ln \frac{{{{\sqrt[3]{{x - 2}}}^{11}}}}{{{{\sqrt[3]{{x + 1}}}^5}}} + C$