Find indefinite integral $\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$ $$\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$$
$$\frac {1}{x^2\sqrt{1-x^2}}+2\int \frac{1}{x^3\sqrt{1-x^2}}\, \mathrm{d}x$$
I try to integrate by parts method, but its doesnt want to be solved. I try to substitue $x=\sin u \mathrm{d}x=\cos u$ but failed somewhere

Let $u=\arcsin x$ and $dv=\frac{dx}{x^2}$. Then we will get (see Michael Burr answer. He used partial fractions and mine is trigo substitution)
$$\int \frac{\arcsin x}{x^2}dx=-\frac{\arcsin x}{x}+\int\frac{dx}{x\sqrt{1-x^2}}.$$
Let $x=\sin \theta$. Then $dx=\cos\theta\ d\theta$ and $$\sqrt{1-x^2}=\cos\theta.$$ Thus,
\begin{align}
\int \frac{\arcsin x}{x^2}dx&=-\frac{\arcsin x}{x}+\int\frac{\cos\theta\ d\theta}{\sin\theta\cos\theta}\\
&=-\frac{\arcsin x}{x}+\int \csc\theta\ d\theta\\
&=-\frac{\arcsin x}{x}-\ln|\csc\theta+\cot\theta|+C\\
&=-\frac{\arcsin x}{x}-\ln\left| \frac{1+\sqrt{1-x^2}}{x}\right|+C\\
&=-\frac{\arcsin x}{x}-\ln\left|1+\sqrt{1-x^2}\right|+\ln|x|+C
\end{align}
