Answer
$$\int \frac{1}{x^{3}\sqrt{x^{2}-1}} dx=\frac{sec^{-1}x}{2}+\frac{\sqrt{x^{2}-1}}{2x^{2}}+C$$
Work Step by Step
$$Let \,x=sec\,t,\,so\,dx= tan\,t\,sec\,t\,dt$$
$$sin\,t=\frac{\sqrt{x^{2}-1}}{x},\,cos\,t=\frac{1}{x}$$
$$\int \frac{1}{x^{3}\sqrt{x^{2}-1}} dx =\int \frac{tan\,t\,sec\,t}{sec^{3}t\,tant\,t} dt$$
$$ = \int cos^{2}t\, dt=\int \frac{1+cos\,2t}{2}dt$$
$$=\frac{t}{2}+\frac{sin2t}{4}+C$$
$$=\frac{t}{2}+\frac{sint\,cost}{2}+C$$
$$=\frac{sec^{-1}x}{2}+\frac{\sqrt{x^{2}-1}}{2x^{2}}+C$$
