Answer
$$\frac{{\sqrt {{x^2} - 1} }}{x} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} - 1} }}} \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}x = \sec \theta,{\text{ }}dx = \sec \theta \tan \theta d\theta,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2} \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} - 1} }}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {\sec \theta } \right)}^2}\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}} \cr
& = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\sec }^2}\theta \sqrt {{{\sec }^2}\theta - 1} }}} \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\sec }^2}\theta \sqrt {{{\tan }^2}\theta } }}} \cr
& {\text{simplifying}} \cr
& = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\sec }^2}\theta \tan \theta }}} \cr
& = \int {\frac{1}{{\sec \theta }}d\theta } \cr
& = \int {\cos \theta d\theta } \cr
& {\text{integrate}} \cr
& = \sin \theta + C \cr
& {\text{write in terms of }}x,{\text{ sin}}\theta = \frac{{\sqrt {{x^2} - 1} }}{x} \cr
& = \frac{{\sqrt {{x^2} - 1} }}{x} + C \cr} $$
