Answer
$= - \frac{3}{4}(4-x^{2})^{\frac{2}{3}} +C $
Work Step by Step
$\int \frac{x}{\sqrt[3] {4-x^{2}}} $ $dx$
Let $u = 4-x^{2}$
$u' = - 2x$
$\frac{du}{dx} = -2x$
$\frac{du}{-2} = x$
$-\frac{1}{2} \int (u)^{-\frac{1}{3}} du $
$= (-\frac{1}{2})(\frac{u^{\frac{2}{3}}}{\frac{2}{3}}) + C$
$ = (-\frac{1}{2})(\frac{3u^{\frac{2}{3}}}{2}) + C$
$= -(\frac{3u^{\frac{2}{3}}}{4}) + C$
$= - \frac{3}{4}u^{\frac{2}{3}} +C $
$= - \frac{3}{4}(4-x^{2})^{\frac{2}{3}} +C $
