Answer
$$
\int(\arcsin x)^{2} d x=x(\arcsin x)^{2}+2 \sqrt{1-x^{2}} \arcsin x-2 x+C
$$
Work Step by Step
$$
\int(\arcsin x)^{2} d x
$$
Integrate by parts twice, first with
$$
\left[\begin{array}{ll}{u=(\arcsin x)^{2},} & {d v=d x} \\ {d u=2\arcsin x \frac{d x}{\sqrt 1-x^{2}} } & {v= x}\end{array}\right]
$$
$$
\int(\arcsin x)^{2} d x=
x(\arcsin x)^{2}-\int 2 x \arcsin x\left(\frac{d x}{\sqrt{1-x^{2}}}\right)
$$
Now let
$$
\left[\begin{array}{ll}{U=\arcsin x ,} & {dV=\left(\frac{x d x}{\sqrt{1-x^{2}}}\right)} \\ {d U=\left(\frac{d x}{\sqrt{1-x^{2}}}\right)} & {V=-\sqrt{1-x^{2}} }\end{array}\right]
$$
So
$$
\begin{aligned}
\int(\arcsin x)^{2} d x &=x(\arcsin x)^{2}-\int 2 x \arcsin x\left(\frac{d x}{\sqrt{1-x^{2}}}\right)\\
&=x(\arcsin x)^{2}-2\left[\arcsin x(-\sqrt{1-x^{2}})+\int d x\right]\\
&= x(\arcsin x)^{2}+2 \sqrt{1-x^{2}} \arcsin x-2 x+C
\end{aligned}
$$
where $C$ is constant.
