Answer
$\frac{1}{5}(x^2+1)^{5/2}-\frac{1}{3}(x^2+1)^{3/2}+C$
Work Step by Step
let u$=x^2+1$
then du=2x dx and $\frac{1}{2}du=xdx$
and $x^2=i-1$
thus
=$\int x^3\sqrt{x^2+1}dx$
=$\int (u-1)\frac{1}{2}\sqrt{u}du$
=$\frac{1}{2}\int (u^{3/2}-\frac{2}{3}u^{1/2})du$
=$\frac{1}{2}[\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}]+C$
=$\frac{1}{5}u^{5/2}-\frac{1}{3}u^{3/2}+C$
=$\frac{1}{5}(x^2+1)^{5/2}-\frac{1}{3}(x^2+1)^{3/2}+C$
