Answer
$$\frac{1}{2} \left[ \sec^{-1}x+ \frac{\sqrt{x^2-1}}{x^2}\right]+c$$
Work Step by Step
Given $$ \int \frac{dx}{x^3\sqrt{x^2-1}}$$
Let $x=\sec u\ \ \Rightarrow \ dx=\sec u \tan udu$, then
\begin{align*}
\int \frac{dx}{x^3\sqrt{x^2-1}}&= \int \frac{ \sec u \tan udu}{\sec^3u\sqrt{\sec^2u-1}}\\
&= \int \frac{ \sec u \tan udu}{\sec^3u\tan u}\\
&= \int \frac{ du}{\sec^2u }\\
&= \int \cos^2u du\\
&=\frac{1}{2} \int(1+ \cos 2u)du\\
&=\frac{1}{2} [u+\frac{1}{2}\sin 2u]+c\\
&=\frac{1}{2} [u+ \sin u\cos u]+c\\
&=\frac{1}{2} \left[ \sec^{-1}x+ \frac{\sqrt{x^2-1}}{x^2}\right]+c
\end{align*}
