Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$

I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)

I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) dt$$
And then evaluating it by parts, which made it very complicated.

So how do I go about solving this?
Let $t= \sqrt{\frac{x}{1-x}}$. Then,

$$I = \int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx
= -\int \arcsin t\&gt; d\left(\frac{1 }{1+t^2}\right) $$

Integrate by parts,

$$I= -\frac{\arcsin t}{1+t^2}+\int \frac{dt}{(1+t^2)\sqrt{1-t^2}} $$

Let $u^2=\frac{2t^2}{1-t^2}$. Then, 
$$\int \frac{dt}{(1+t^2)\sqrt{1-t^2}} = \frac1{\sqrt2}\int \frac{du}{u^2+1} 
= \frac1{\sqrt2}\arctan u$$

Thus,

$$I= -\frac{\arcsin t}{1+t^2}+ \frac1{\sqrt2}\arctan \left(\frac{\sqrt2 t}{\sqrt{1-t^2}}\right)+C$$
$$=-(1-x) \arcsin\sqrt{\frac{x}{1-x}}+\frac{1}{\sqrt{2}}\arctan\sqrt{\frac{2x}{1-2x}}+ C$$
This is what you should get. I am very sorry I thought in the denominator its not arcSin, my apologies!!



Please go through this, and let me know if you have concerns or questions.