Đáp án:
\[\int {\frac{{{x^3}}}{{\sqrt {2 - {x^2}} }}dx} = - 2\sqrt {2 - {x^2}} + \frac{{{{\sqrt {2 - {x^2}} }^3}}}{3} + C\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
t = 2 - {x^2} \Rightarrow dt = \left( {2 - {x^2}} \right)'dx = - 2xdx\\
t = 2 - {x^2} \Leftrightarrow {x^2} = 2 - t\\
\int {\frac{{{x^3}}}{{\sqrt {2 - {x^2}} }}dx} = \int {\frac{{{x^2}.\left( { - 2xdx} \right)}}{{\left( { - 2} \right).\sqrt {2 - {x^2}} }}} \\
= \frac{{ - 1}}{2}.\int {\frac{{\left( {2 - t} \right)dt}}{{\sqrt t }}} \\
= - \frac{1}{2}.\int {\left( {\frac{2}{{\sqrt t }} - \sqrt t } \right)dt} \\
= \frac{{ - 1}}{2}.\left( {2\int {{t^{\frac{{ - 1}}{2}}}} dt - \int {{t^{\frac{1}{2}}}dt} } \right)\\
= \frac{{ - 1}}{2}.\left( {2.\frac{{{t^{ - \frac{1}{2} + 1}}}}{{\frac{{ - 1}}{2} + 1}} - \frac{{{t^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}}} \right) + C\\
= \frac{{ - 1}}{2}.\left( {4{t^{\frac{1}{2}}} - \frac{2}{3}{t^{\frac{3}{2}}}} \right) + C\\
= - 2\sqrt t + \frac{{\sqrt {{t^3}} }}{3} + C\\
= - 2\sqrt {2 - {x^2}} + \frac{{{{\sqrt {2 - {x^2}} }^3}}}{3} + C
\end{array}\)
