Answer
$-\dfrac{1}{3}(\dfrac{x}{\sqrt{x^2-1}})^3+C$
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^2 \theta (\sec \theta \tan \theta)d \theta }{(tan^5 \theta)} =\int \dfrac{\cos \theta }{\sin^4 \theta } d \theta $
Let us substitute $u=\sin \theta \implies du =\cos \theta d \theta$
we have $=\dfrac{1}{u^4} du$
Now, integrate .
$\dfrac{-u^3}{3}+C=\dfrac{-1}{3\sin^3 \theta}+C=-\dfrac{1}{3}\csc^3 \theta +C$
and plug in $\tan \theta=\sin \theta /\cos \theta =\sqrt {x^2-1} \implies \csc \theta=\dfrac{1}{\sin \theta} =\dfrac{x}{\sqrt {x^2 -1}}$:
$-\dfrac{1}{3}\csc^3 \theta +C=-\dfrac{1}{3}(\dfrac{x}{\sqrt{x^2-1}})^3+C$
