Đáp án: $\frac{1}{5}\ln \left( {x - 3} \right) - \frac{6}{5}\ln \left( {x + 2} \right) + C$
Giải thích các bước giải:
$\begin{array}{l}
\frac{{4 - x}}{{{x^2} - x - 6}} = \frac{{4 - x}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
= \frac{{a\left( {x + 2} \right) + b\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
= \frac{{\left( {a + b} \right)x + 2a - 3b}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\\
\Rightarrow \left\{ \begin{array}{l}
a + b = - 1\\
2a - 3b = 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = \frac{1}{5}\\
b = - \frac{6}{5}
\end{array} \right.\\
\Rightarrow \frac{{4 - x}}{{{x^2} - x - 6}} = \frac{1}{{5\left( {x - 3} \right)}} - \frac{6}{{5\left( {x - 2} \right)}}\\
\Rightarrow \int {\frac{{4 - x}}{{{x^2} - x - 6}}} dx = \int {\frac{1}{{5\left( {x - 3} \right)}} - \frac{6}{{5\left( {x - 2} \right)}}dx} \\
= \frac{1}{5}\ln \left( {x - 3} \right) - \frac{6}{5}\ln \left( {x + 2} \right) + C
\end{array}$
