Answer
$=2\tan^{-1}\sqrt{x-1}+C$
Work Step by Step
$\displaystyle \int\frac{dx}{x\sqrt{x-1}}=\quad \left[\begin{array}{cc}
{t=\sqrt{x-1}},& {x=t^{2}+1}\\
{t^{2}=x-1},\quad & {dx=2tdt}\end{array}\right]$
$=\displaystyle \int\frac{2t}{t\left(t^{2}+1\right)}dt$
$=2\displaystyle \int\frac{1}{t^{2}+1}dt$
$=2\tan^{-1}t+C$
$=2\tan^{-1}\sqrt{x-1}+C$
