Đáp án:
$\begin{array}{l}
y = \frac{{3{x^2} + 3x + 3}}{{{x^3} - 3x + 2}} = \frac{{3{x^2} + 3x + 3}}{{{{\left( {x - 1} \right)}^2}.\left( {x + 2} \right)}}\\
y = \frac{A}{{{{\left( {x - 1} \right)}^2}}} + \frac{B}{{x - 1}} + \frac{C}{{x + 2}}\\
= \frac{{A\left( {x + 2} \right) + B\left( {x - 1} \right)\left( {x + 2} \right) + C{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}\\
= \frac{{A.x + 2A + B{x^2} + Bx - 2B + C{x^2} - 2C.x + C}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}\\
= \frac{{\left( {B + C} \right){x^2} + \left( {A + B - 2C} \right).x + \left( {2A - 2B + C} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}\\
\Rightarrow \left( {B + C} \right){x^2} + \left( {A + B - 2C} \right).x + \left( {2A - 2B + C} \right)\\
= 3{x^2} + 3x + 3\\
\Rightarrow \left\{ \begin{array}{l}
B + C = 3\\
A + B - 2C = 3\\
2A - 2B + C = 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
A = 3\\
B = 2\\
C = 1
\end{array} \right.\\
\Rightarrow y = \frac{3}{{{{\left( {x - 1} \right)}^2}}} + \frac{2}{{x - 1}} + \frac{1}{{x + 2}}\\
b)\int y = \int {\frac{3}{{{{\left( {x - 1} \right)}^2}}} + \frac{2}{{x - 1}} + \frac{1}{{x + 2}}} dx\\
= \frac{{ - 3}}{{x - 1}} + 2\ln \left| {x - 1} \right| + \ln \left| {x + 2} \right| + C\\
c)F\left( 2 \right) = 0\\
\Rightarrow \frac{{ - 3}}{{2 - 1}} + 2\ln \left| {2 - 1} \right| + \ln \left| {2 + 2} \right| + C = 0\\
\Rightarrow - 3 + \ln 4 + C = 0\\
\Rightarrow C = 3 - 2\ln 2\\
F\left( x \right) = \frac{{ - 3}}{{x - 1}} + 2\ln \left| {x - 1} \right| + \ln \left| {x + 2} \right| + 3 - 2\ln 2
\end{array}$
