Answer
$${\text{False}}$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{x^2} - 1} }}{x}} dx \cr
& {\text{Let }}x = \sec \theta ,{\text{ }}dx = \sec \theta \tan \theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{\sqrt {{x^2} - 1} }}{x}} dx = \int {\frac{{\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}{{\sec \theta }}} \left( {\sec \theta \tan \theta } \right)d\theta \cr
& {\text{Simplifying}} \cr
& = \int {\sqrt {{{\sec }^2}\theta - 1} } \left( {\tan \theta } \right)d\theta \cr
& = \int {\sqrt {{{\tan }^2}\theta } } \left( {\tan \theta } \right)d\theta \cr
& = \int {{{\tan }^2}\theta } d\theta \cr
& {\text{The statement is False}} \cr} $$
