Đáp án:
$\begin{array}{l}
\int {\frac{{{x^3}}}{{{x^2} + 1}}dx} \\
= \int {\frac{1}{2}.\frac{{{x^2}}}{{{x^2} + 1}}.2xdx} \\
= \frac{1}{2}\int {\frac{{{x^2}}}{{{x^2} + 1}}d\left( {{x^2} + 1} \right)} \\
= \frac{1}{2}\int {1 - \frac{1}{{{x^2} + 1}}} d\left( {{x^2} + 1} \right)\\
= \frac{1}{2}\left( {{x^2} + 1 - \ln \left( {{x^2} + 1} \right)} \right)\\
= \frac{{{x^2} + 1 - \ln \left( {{x^2} + 1} \right)}}{2}
\end{array}$
