Answer
$\displaystyle \frac{2x^{4}-1}{8}\sin^{-1}(x^{2})+\frac{x^{2}\sqrt{1-x^{4}}}{8}+C$
Work Step by Step
$\displaystyle \int x^{3}\arcsin(x^{2})dx=\quad$substitute $\left[\begin{array}{ll}
u=x^{2} & \\
du=2xdx & xdx=\frac{1}{2}du
\end{array}\right]$
$=\displaystyle \frac{1}{2}\int u\arcsin(u)du=$
Table of integrals:
$\color{blue}{ 90. \quad \displaystyle \int u\sin^{-1}udu=\frac{2u^{2}-1}{4}\sin^{-1}u+\frac{u\sqrt{1-u^{2}}}{4}+C }$
$=\displaystyle \frac{2u^{2}-1}{8}\sin^{-1}u+\frac{u\sqrt{1-u^{2}}}{8}+C$
... bring back x...
$=\displaystyle \frac{2x^{4}-1}{8}\sin^{-1}(x^{2})+\frac{x^{2}\sqrt{1-x^{4}}}{8}+C$
