Đáp án:
$\begin{array}{l}
I = \int {\frac{{{x^3}}}{{\sqrt {2 - {x^2}} }}dx} \\
= \int {\frac{{{x^2}.2x}}{{2\sqrt {2 - {x^2}} }}dx} \\
= \int {\frac{{{x^2}}}{{2\sqrt {2 - {x^2}} }}d\left( {{x^2}} \right)} \\
Đặt\,\sqrt {2 - {x^2}} = u\\
\Rightarrow 2 - {x^2} = {u^2}\\
\Rightarrow \left\{ \begin{array}{l}
d\left( {{x^2}} \right) = - 2udu\\
{x^2} = 2 - {u^2}
\end{array} \right.\\
\Rightarrow I = \int {\frac{{2 - {u^2}}}{{2u}}.\left( { - 2u} \right)du} \\
= \int {\left( {{u^2} - 2} \right)du} \\
= \frac{1}{3}{u^3} - 2u + C\\
= \frac{1}{3}.\left( {2 - {x^2}} \right)\sqrt {2 - {x^2}} - 2\sqrt {2 - {x^2}} + C
\end{array}$
