Answer
$x(\arcsin x)^{2}+2\sqrt{1-x^{2}}\arcsin x-2x+C$
Work Step by Step
Integration by parts:
$\displaystyle \int udv=uv-\int vdu$
The idea is to choose a relatively easy $dv$ to integrate, and
a u whose $u'$ does not complicate matters (best case: makes thing simpler).
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Let $u=(\arcsin x)^{2},\qquad dv=dx$
Then,
$du=2\displaystyle \arcsin x\cdot\frac{1}{\sqrt{1-x^{2}}}, \qquad v=x$
$I=\displaystyle \int(\arcsin x)^{2}dx=uv-\int vdu$
$=x(\displaystyle \arcsin x)^{2}-2\int\arcsin x\cdot\frac{x}{\sqrt{1-x^{2}}}dx$
It might not look like it, but we have a "simpler" integrand.
By parts again, $\left[\begin{array}{ll}
u=\arcsin x & dv=\frac{x}{\sqrt{1-x^{2}}}\\
du=\frac{1}{\sqrt{1-x^{2}}} & v= -\sqrt{1-x^{2}}
\end{array}\right],$
$($ for $\displaystyle \int\frac{xdx}{\sqrt{1-x^{2}}}$= substitute $\left[\begin{array}{l}
t=1-x^{2}\\
dt=-2xdx
\end{array}\right]$
= $-\displaystyle \frac{1}{2}\int t^{-1/2}dt=-\frac{1}{2}(\frac{t^{1/2}}{1/2})=-t^{1/2})$
$I=x(\displaystyle \arcsin x)^{2}-2(-\sqrt{1-x^{2}}\arcsin x-\int-1dx)$
$=x(\arcsin x)^{2}+2\sqrt{1-x^{2}}\arcsin x-2x+C$