Tìm nguyên hàm $\int\ {\frac{x^3}{\sqrt{2- x^2}}} \, dx$

Đáp án:

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