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?

$$I=\int \sin^{-1} \sqrt{\frac{x}{1-x}} dx$$
Let $$\frac{x}{1-x}=t^2 \implies x=\frac{t^2}{1+t^2} \implies dx=\frac{2t}{(1+t^2)^2}$$
Then $$I=\int\frac{2t \sin^{-1}t}{(1+t^2)^2} ~dt $$
Integrate it by parts taking $\sin^{-1} t$ as the first function.
Then $$I=\sin^{-1} t \int \frac{ 2t dt}{(1+t^2)^2}=\frac{-\sin^{-1}t}{1+t^2}+\int \frac{dt}{(1+t^2)\sqrt{1-t^2}}=f(t)+I_1$$
Let $t=1/u \implies du=- dt/t^2$ and then let $v=u^2$, then
$$I_1=-\int \frac{u du}{(1+u^2)\sqrt{u^2-1}}=-\int \frac{v dv}{2(1+v)\sqrt{v-1}}+C$$
Next take $v=w^2 \implies dv=2wdw$, then
$$I_1=-\int \frac{dw}{2+w^2}=-\frac{1}{\sqrt{2}} \tan^{-1} \frac{w}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\tan^{-1} \frac{1-2x}{2x}$$
Finally, $$I=(x-1)\sin^{-1} \sqrt{\frac{x}{1-x}}-\frac{1}{\sqrt{2}}\tan^{-1} \frac{1-2x}{2x}$$
