Answer
$$
\int \frac{d x}{x \sqrt{x-1}} =2 \tan ^{-1} \sqrt{x-1}+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int \frac{d x}{x \sqrt{x-1}}
$$
By substituting
$$
\left[\begin{array}{c}
u=\sqrt{x-1}, x=u^{2}+1 \\
u^{2}=x-1, d x=2 u d u
\end{array}\right]
$$
we get:
$$
\begin{aligned}
\int \frac{d x}{x \sqrt{x-1}} &=\int \frac{2 u}{u\left(u^{2}+1\right)} d u \\
&=2 \int \frac{1}{u^{2}+1} d u \\
& =2 \tan ^{-1} u+C \\
&=2 \tan ^{-1} \sqrt{x-1}+C
\end{aligned}
$$
where $C$ is an arbitrary constant.
