<s>[INST] 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?
 [/INST] Substitute 
$$u = \sqrt{\frac{x}{1-x}} \implies \frac{du}{dx} = \frac{\sqrt{x}}{(1-x)^{\frac{3}{2}}} + \frac{1}{2\sqrt{1-x}\sqrt{x}}$$
Then the integral become:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du$$
Integrating by parts $\left(f = \arcsin(u), g'= \frac{u}{(u^2 + 1)^2} \implies f'= \frac{1}{\sqrt{1 - u^2}}, g = -\frac{1}{2(u^2 + 1)}\right)$:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du = -\frac{\arcsin u}{2(u^2+1)} - \int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du \text{ (1)}$$
We will now solve this integral first:
$$\int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du \text{ (2)}$$
Substitute $v = \arcsin(u) \implies du = \cos(v) \,dv$, then:
$$(2) \implies \int \frac{\cos(v)}{2\sqrt{1-\sin(v)^2}(\sin(v)^2+1)} \,dv$$
$$= \frac{1}{2} \int \frac{1}{\sin(v)^2+1} \,dv$$
$$= \frac{1}{2} \int \sec^2(v) \cdot \frac{1}{2\tan(v)^2+1} \,dv \text{ (3)}$$
Substitute $w = \sqrt{2}\tan(v) \implies \frac{dw}{dv} = \sqrt{2}\sec^2(v)$, then:
$$(3) \implies \frac{1}{2\sqrt{2}} \int \frac{1}{w^2+1} \,dw = \frac{1}{2\sqrt{2}} \arctan(w)$$
Undo substitution, we get:
$$\int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du = \frac{\arctan(\frac{\sqrt{2}u}{\sqrt{1-u^2}})}{2\sqrt{2}}$$ 
Plug it back into $(1)$ and undo substitution to get the answer.
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