Answer
$\dfrac{\sqrt{x^2 -1}}{x}+C$
Work Step by Step
Let us consider $x = \sec \theta $
and $dx= \sec\theta \tan \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{ \sec \theta \tan \theta }{ \sec^2 \theta\sqrt{\sec^2 \theta -1}} d \theta = \int cos \theta d \theta$
Now, integrate and plug in $\sec \theta =x$:
$ \sin \theta+C=\dfrac{\sqrt{x^2 -1}}{x}+C$
