How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$? I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$
I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.

Substitute $u=\sqrt{x-1}$. Then $2du=\dfrac{1}{\sqrt{x-1}}dx$ and $x=u^2+1$. 
Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\int \dfrac{2}{(u^2+1)^2}du.$$
Put $U=\dfrac{1}{u^{2}+1}$ and $dV=du$. Then integration by parts,
$$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{u^{2}+1}+\int \dfrac{2u^2}{(u^2+1)^2}du\\=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{u^{2}+1}{(u^{2}+1)^2}-\dfrac{1}{(u^{2}+1)^2}\right)du\\
=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{1}{(u^{2}+1)}-\dfrac{1}{(u^{2}+1)^2}\right)du $$
Hence $$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\int \dfrac{1}{(u^{2}+1)}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\tan^{-1}(u).$$
Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\dfrac{\sqrt{x-1}}{x}+\tan^{-1}(\sqrt{x-1}).$$
