Help with integral with $\arcsin x$.

Question

$$\int \frac{(1+x^2)\arcsin x}{x^2\sqrt{1-x^2}}dx$$ I saw that $$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$$ and I tried to solve it "by parts"

Answer 1

$$ I = \int \frac{x dx}{\sin^2 x} $$

$$ u = x, v = \int \frac{dx}{\sin^2 x} $$

$$ I = uv - \int vdu $$

$$ v = -\,\mathrm{ctg}\,x $$

$$ I = -x\,\mathrm{ctg}\,x + \int \mathrm{ctg}\,x dx $$

$$ \int \mathrm{ctg}\,x dx = \ln |\sin x| + C $$

Answer 2

Hint Try $x = \sin u, dx = \cos u du$ to get $$ \int \frac{(1+x^2)\arcsin x}{x^2\sqrt{1-x^2}}dx = \int \frac{(1+\sin^2 u) u \cos u du}{\sin^2 u\sqrt{1-\sin^2 u}} = \int \left(\csc^2 u + 1\right) u du $$ which splits into 2 standard integrals.