Answer
$${\sec ^{ - 1}}x + \frac{{\sqrt {{x^2} - 1} }}{{{x^2}}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{2dx}}{{{x^3}\sqrt {{x^2} - 1} }}} \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}x = \sec \theta,{\text{ }}dx = \sec \theta \tan \theta d\theta,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2} \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{2dx}}{{{x^3}\sqrt {{x^2} - 1} }}} = \int {\frac{{2\left( {\sec \theta \tan \theta } \right)d\theta }}{{{{\left( {\sec \theta } \right)}^3}\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}} \cr
& = \int {\frac{{2\sec \theta \tan \theta d\theta }}{{{{\sec }^3}\theta \sqrt {{{\sec }^2}\theta - 1} }}} \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{2\sec \theta \tan \theta d\theta }}{{{{\sec }^3}\theta \sqrt {{{\tan }^2}\theta } }}} \cr
& {\text{simplifying}} \cr
& = \int {\frac{{2\sec \theta \tan \theta d\theta }}{{{{\sec }^3}\theta \tan \theta }}} \cr
& = \int {\frac{2}{{{{\sec }^2}\theta }}d\theta } \cr
& = 2\int {{{\cos }^2}\theta d\theta } \cr
& = 2\int {\frac{1}{2}\left( {1 + \cos 2\theta } \right)} d\theta \cr
& = \int {\left( {1 + \cos 2\theta } \right)} d\theta \cr
& {\text{integrate}} \cr
& = \theta + \frac{1}{2}\sin 2\theta + C \cr
& = \theta + \sin \theta \cos \theta + C \cr
& {\text{write in terms of }}x,{\text{ sin}}\theta = \frac{{\sqrt {{x^2} - 1} }}{x}{\text{ and cos}}\theta = \frac{1}{x} \cr
& = {\sec ^{ - 1}}x + \left( {\frac{{\sqrt {{x^2} - 1} }}{x}} \right)\left( {\frac{1}{x}} \right) + C \cr
& = {\sec ^{ - 1}}x + \frac{{\sqrt {{x^2} - 1} }}{{{x^2}}} + C \cr} $$
