Congratulations to MarkFL, lfdahl, Opalg and Theia for their correct solutions.
MarkFL's solution:
[sp] Let:
$$I=\int \arcsin^2(x)\,dx$$
Using IBP, I would state:
$$u=\arcsin(x)\implies du=\frac{1}{\sqrt{1-x^2}}\,dx$$
$$dv=\arcsin(x)\,dx$$
To find \(v\), let's use IBP where:
$$r=\arcsin(x)\implies dr=\frac{1}{\sqrt{1-x^2}}\,dx$$
$$ds=dx\implies s=x$$
Hence:
$$v=x\arcsin(x)-\int \frac{x}{\sqrt{1-x^2}}\,dx=x\arcsin(x)+\sqrt{1-x^2}$$
And so we have:
$$I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-\int \left(x\arcsin(x)+\sqrt{1-x^2}\right)\frac{1}{\sqrt{1-x^2}}\,dx$$
$$I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-\int \frac{x}{\sqrt{1-x^2}}\arcsin(x)+1\,dx$$
$$I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-x-\int \frac{x}{\sqrt{1-x^2}}\arcsin(x)\,dx$$
On the remaining integral, use IBP where:
$$u=\arcsin(x)\implies du=\frac{1}{\sqrt{1-x^2}}\,dx$$
$$dv= \frac{x}{\sqrt{1-x^2}}\,dx\implies v=-\sqrt{1-x^2}$$
And so we have:
$$I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-x-\left(-\sqrt{1-x^2}\arcsin(x)+\int \sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}\,dx\right)$$
And so we conclude:
$$I=\int \arcsin^2(x)\,dx=\arcsin(x)\left(x\arcsin(x)+2\sqrt{1-x^2}\right)-2x+C$$[/sp]
Opalg's solution:
[sp] Let $u = \arcsin(x)$. Then $x = \sin u$, $dx = \cos u\,du$, and $$\int \arcsin^2(x)\,dx = \int u^2\cos u\,du$$. Integrate by parts twice to get $$\begin{aligned} \int u^2\cos u\,du &= u^2\sin u - \int2u\sin u\,du \\ &= u^2\sin u + 2u\cos u - \int2\cos u\,du \\ &= u^2\sin u + 2u\cos u - 2\sin u + \text{const.} \end{aligned}$$ Now substitute back in terms of $x$ to get $$\int \arcsin^2(x)\,dx = x(\arcsin^2(x) - 2) + 2\sqrt{1-x^2}\arcsin(x) + \text{const.}$$
(Finally, differentiate that result to check that the derivative is indeed $\arcsin^2(x)$.)[/sp]