Answer
$= \frac{1}{3} (x^{2} - 36)^{\frac{3}{2}} + C$
Work Step by Step
$\int x\sqrt {x^{2} - 36}$ $dx$
Let $u = x^{2} - 36$
$u' = 2x$
$\frac{du}{dx} = 2x$
$\frac{du}{2} = x$
$\frac{1}{2} \int u^{\frac{1}{2}} du$
$= \frac{1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} +C$
$= \frac{1}{2} \frac{2u^{\frac{3}{2}}}{3} + C$
$= \frac{2u^{\frac{3}{2}}}{6} + C$
$= \frac{u^{\frac{3}{2}}}{3} + C$
$= \frac{1}{3} u^{\frac{3}{2}} + C$
$= \frac{1}{3} (x^{2} - 36)^{\frac{3}{2}} + C$
