Đáp án:
$\begin{array}{l}
\int\limits_0^1 {\frac{{{x^3}}}{{{x^4} + 3{x^2} + 2}}dx} \\
= \int\limits_0^1 {\frac{1}{2}.\frac{{{x^2}}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}.2xdx} \\
= \int\limits_0^1 {\frac{{{x^2}}}{{2\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}d{x^2}} \\
= \int\limits_0^1 {\frac{u}{{2\left( {u + 1} \right)\left( {u + 2} \right)}}du} \\
Do:\frac{u}{{\left( {u + 1} \right)\left( {u + 2} \right)}} = \frac{a}{{u + 1}} + \frac{b}{{u + 2}}\\
= \frac{{\left( {a + b} \right)u + 2a + b}}{{\left( {u + 1} \right)\left( {u + 2} \right)}}\\
\Rightarrow \left\{ \begin{array}{l}
a + b = 1\\
2a + b = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = - 1\\
b = 2
\end{array} \right.\\
\Rightarrow I = \frac{1}{2}.\int\limits_0^1 {\frac{{ - 1}}{{u + 1}} + \frac{2}{{u + 2}}du} \\
= \left( {\ln \left| {u + 2} \right| - \frac{1}{2}.\ln \left| {u + 1} \right|} \right)_0^1\\
= \ln 3 - \frac{3}{2}\ln 2
\end{array}$
