Đáp án:
$\begin{array}{l}
A = \dfrac{{3{x^3} - x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= 3\left( {{x^3} - 3{x^2} + 2x} \right) + 9{x^2} - 27x + 18\\
+ 20x - 19\\
= \left( {3x + 9} \right)\left( {{x^2} - 3x + 2} \right) + 20x - 19\\
= \left( {3x + 9} \right).\left( {x - 1} \right)\left( {x - 2} \right) + 20x - 19\\
\Rightarrow A = \dfrac{{3{x^3} - x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \dfrac{{\left( {3x + 9} \right)\left( {x - 1} \right)\left( {x - 2} \right) + 20x - 19}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= 3x + 9 + \dfrac{{20x - 19}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
Đặt:\dfrac{{20x - 19}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{a.\left( {x - 2} \right) + b\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
\Rightarrow 20x - 19 = a.\left( {x - 2} \right) + b\left( {x - 1} \right)\\
= a.x - 2a + b.x - b\\
= \left( {a + b} \right).x - 2a - b\\
\Rightarrow \left\{ \begin{array}{l}
a + b = 20\\
- 2a - b = 19
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = - 39\\
b = 59
\end{array} \right.\\
\Rightarrow \dfrac{{20x - 19}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \dfrac{{ - 39\left( {x - 2} \right) + 59\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \dfrac{{59}}{{x - 2}} - \dfrac{{39}}{{x - 1}}\\
\Rightarrow \int {\dfrac{{3{x^3} - x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}dx} \\
= \int {3x + 9 + \dfrac{{59}}{{x - 2}} - \dfrac{{39}}{{x - 1}}dx} \\
= \dfrac{3}{2}{x^2} + 9x + 59\ln \left| {x - 2} \right| - 39\left| {x - 1} \right| + C
\end{array}$
