Answer
$\sqrt{x^2-1}+C$
Work Step by Step
Consider $x=\sec \theta \implies dx=\sec \theta \tan \theta d \theta $
Now, the given integral can be written as:
$\int\dfrac{\sec^2 \theta \tan \theta}{ \sqrt{\tan^2 \theta}} d\theta= \int \sec^2 \theta$
or, $= \tan d \theta+C$
or, $=\sqrt{x^2-1}+C$
