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.

\begin{align}
\int \dfrac{1}{x^2 \sqrt{x-1}}dx=&\int \frac{1}{\sqrt{x-1}}d\left(\frac{x-1}x\right)\\
= &\ \frac{\sqrt{x-1}}x +\int \frac{ d(\sqrt{x-1})}{x}
= \frac{\sqrt{x-1}}x +\tan^{-1} \sqrt{x-1}
\end{align}
