solve this integral using integral by parts:
$$\int\sin^{-1}(x) \frac{x}{\sqrt{(1-x^2)^2}}\,dx$$
I used substitution : $\sin^{-1}(x)=t , (so,\sin t=x$), $dt=\frac{1}{\sqrt{1-x^2}}\,dx$
$$\int t \frac{\sin t}{\sqrt{1-\sin^2t}}\,dt= \int t\tan t\, dt$$ now I need some help to calculate: $\int t\tan t\, dt$
There is no "elementary" solution. Once the problem is reduced to the computation of $\int t\tan(t)\,dt$, integration by parts gives that this problem is equivalent to the computation of $\int \log\cos t\,dt$. By the Fourier series of $\log\cos$ we have
$$ \int\log\cos(t)\,dt =C+t\log(2)+\sum_{k\geq 1}\frac{(-1)^k}{2k^2}\sin(2kt)$$
where
$$ \sum_{k\geq 1}\frac{(-1)^k}{k^2}\sin(2kt) = \text{Im}\sum_{k\geq 1}\frac{e^{ki(2t+\pi)}}{k^2} =\text{Im}\,\text{Li}_2\left(-e^{2it}\right)$$
and there's no way to avoid dilogarithms in the computation of a primitive.
On the other hand there are special values for particular integration ranges, for instance
$$ \int_{0}^{\pi/4}\log\cos(t)\,dt = \frac{K}{2}-\frac{\pi}{4}\log(2) $$
with $K$ being Catalan's constant, i.e. $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}$.
Comment:
$$\int t. tan (t) dt=-t Ln Cos(t)+ \int Ln Cos(t) dt$$
Wolfram alfa says:
$$\int Ln Cos(t) dt=\frac{1}{2}i(Li_2(-e^{2it})+t(t+2i Ln(1+2^{2it})-2it Ln(Cos(t)+ constant$$
$$\int\sin^{-1}(x) \frac{x}{\sqrt{(1-x^2)^2}}dx$$ $\sin^{-1}(x)$=$\cos^{-1} \sqrt{1-x^2}$ ,$\sqrt{1-x^2}=u$,$du= \frac{-x}{\sqrt{1-x^2}}$
$$\int \frac{-cos^{-1}(u)}{u}du$$ and we need to calculate this but how?
