Answer
$$ - \frac{x}{{\sqrt {{x^2} - 1} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {{x^2} - 1} \right)}^{3/2}}}}} \cr
& {\text{We set}}{\text{, }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{dx}}{{{{\left( {{x^2} - 1} \right)}^{3/2}}}}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {{{\left( {\sec \theta } \right)}^2} - 1} \right)}^{3/2}}}}} \cr
& = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {{{\sec }^2}\theta - 1} \right)}^{3/2}}}}} \cr
& \cr
& {\text{Use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {{{\tan }^2}\theta } \right)}^{3/2}}}}} \cr
& = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\tan }^3}\theta }}} \cr
& \cr
& {\text{simplifying}} \cr
& = \int {\frac{{\sec \theta }}{{{{\tan }^2}\theta }}d\theta = \int {\left( {\frac{1}{{\cos \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)} d\theta } \cr
& = \int {\frac{{\cos \theta }}{{{{\sin }^2}\theta }}} d\theta \cr
& = \int {{{\left( {\sin \theta } \right)}^{ - 2}}\left( {\cos \theta } \right)} d\theta \cr
& \cr
& {\text{integrate}} \cr
& = \frac{{{{\left( {\sin \theta } \right)}^{ - 1}}}}{{ - 1}} + C \cr
& = - \frac{1}{{\sin \theta }} + C \cr
& = - \csc \theta + C \cr
& \cr
& {\text{write in terms of }}x,{\text{ csc}}\theta = \frac{x}{{\sqrt {{x^2} - 1} }} \cr
& = - \frac{x}{{\sqrt {{x^2} - 1} }} + C \cr} $$
