Find integral $\int\frac{\arcsin(x)}{x^{2}}dt$ Find integral $$\int\frac{\arcsin(x)}{x^{2}}dx$$
what I've done: $$\int\frac{\arcsin(x)}{x^{2}}dx=-\int\arcsin(x)d(\frac{1}{x})=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$ I got stuck with that

Perform integration by parts to get that $\displaystyle \int \frac{\arcsin(x)}{x^2} dx$ = $\displaystyle  \frac{-\arcsin(x)}{x} + \int \frac{1}{x\sqrt{1 -x^2}}dx$.
To solve this latter integral, let $x = \sin(\theta)$. Then $dx = \cos(\theta)d\theta,$ so the integral becomes $\displaystyle \int\frac{\cos(\theta)}{\sin(\theta)\cos(\theta)}d\theta = \int \csc(\theta) d\theta  = -\ln|\csc(\theta) + \cot(\theta)| + C.$
Now since we have $x = \sin(\theta)$, then $\frac{1}{x} = \csc(\theta)$ and $\cot(\theta) = \frac{\sqrt{1 - x^2}}{x}$. 
Putting it all together, we have $\displaystyle \int \frac{\arcsin(x)}{x^2} dx =\displaystyle  \frac{-\arcsin(x)}{x} - \ln|\frac{1+ \sqrt{1 - x^2}}{x}| + C. $
