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