<s>[INST] 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
 [/INST] $\int \frac{\arcsin(x)}{x^2}dx=\int \arcsin(x) d(-\frac{1}{x})=-\frac{1}{x}\arcsin(x)+\int \frac{dx}{x \sqrt{1-x^2}}= -\frac{1}{x}\arcsin(x)+\int \frac{dx}{x^2 \sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)+\int \frac{d(-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)-\int \frac{dy}{\sqrt{y^2-1}}=-\frac{1}{x}\arcsin(x)-\ln(y+\sqrt{y^2-1})=-\frac{1}{x}\arcsin(x)-\ln(\frac{1}{x}+\sqrt{(1/x)^2-1})$
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