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
 A: 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. $
A: Hint. For the rightmost term, try $x=\sin \theta.$
A: Hint: Use $t=\sqrt{1-x^2}$, $x=\sqrt{1-t^2}$, $dx=-\frac{t}{\sqrt{1-t^2}}dx$  so:
$$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int\frac{1}{1-t^2}dt$$
Then partial fraction expansion: $-\frac{1}{1-t^2}=\frac{1}{2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)$
A: $$\int\frac{\arcsin(x)}{x^{2}}dx=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$
Let $$u=\sqrt{1-x^2}$$
Hence
$$\frac{du}{dx}=\frac{-x}{\sqrt{1-x^2}}$$
$$\int \frac{dx}{x\sqrt{1-x^2}}=-\int \frac{du}{x^2}=-\int\frac{du}{1-u^2}=-\tanh^{-1}(u)+C=-\tanh^{-1}\left(\sqrt{1-x^2}\right)+C$$
Hence 
\begin{align}
\int\frac{\arcsin(x)}{x^{2}}dx &=-\frac{\arcsin(x)}{x}-\tanh^{-1}\left( \sqrt{1-x^2}\right)+C \\
&=-\frac{\arcsin(x)}{x}-\frac12\left(\ln\left(1+\sqrt{1-x^2}\right)+\ln\left(1-\sqrt{1-x^2} \right) \right)+C
\end{align}
