Answer
$\sec^{-1} x+\dfrac{\sqrt{x^2 -1}}{x^2}+C$
Work Step by Step
Let us consider $x = \sec \theta $
and $dx= \sec\theta \tan \theta d \theta $ and $tan \theta =\sqrt {x^2-1}$
Now, the given integral becomes:
$\int \dfrac{2 \sec \theta \tan \theta }{ \sec^3 \theta\tan \theta} d \theta =\int 2 \cos^2 \theta d \theta$
$2\cos^2 \theta =1+\cos 2 \theta$
Then, $\int 2 \cos^2 \theta d \theta= \int (1+cos 2 \theta ) d \theta$
Now, integrate and plug in $\sec \theta =x \implies \theta =\sec^{-1} x $ and $\cos \theta =\dfrac{1}{x} $ and $\sin \theta =\dfrac{\sqrt {x^2-1}}{x}$ :
$\theta+\dfrac{1}{2} \sin (2 \theta)+C=\theta+\dfrac{1}{2} (2 \sin \theta \cos \theta)+C=\sec^{-1} x+\dfrac{\sqrt{x^2 -1}}{x^2}+C$
