Answer
$$\ln \left| {x + \sqrt {{x^2} - 1} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} \cr
& {\text{Substitute }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr
& {\text{Use the trigonometric substitution}} \cr
& \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \int {\frac{{\sec \theta \tan \theta }}{{\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}} d\theta \cr
& {\text{Simplify}} \cr
& = \int {\frac{{\sec \theta \tan \theta }}{{\sqrt {{{\sec }^2}\theta - 1} }}} d\theta \cr
& {\text{Where se}}{{\text{c}}^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{\sec \theta \tan \theta }}{{\sqrt {{{\tan }^2}\theta } }}} d\theta \cr
& = \int {\frac{{\sec \theta \tan \theta }}{{\tan \theta }}} d\theta \cr
& = \int {\sec \theta } d\theta \cr
& {\text{Integrate by the formula }}\int {\sec x} dx = \ln \left| {\sec x + \tan x} \right| + C \cr
& = \ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{Writing the answer in terms of }}x,{\text{ we have that }} \cr
& x = \sec \theta ,\,\,\,\,\tan \theta = \sqrt {{x^2} - 1} \cr
& {\text{Then}}{\text{,}} \cr
& = \ln \left| {x + \sqrt {{x^2} - 1} } \right| + C \cr} $$
