Answer
$$\sqrt {{x^2} - 25} - 5{\sec ^{ - 1}}\left( {\frac{x}{5}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{x^2} - 25} }}{x}} dx \cr
& {\text{Using substitution }}x = 5\sec \theta \cr
& x = 5\sec \theta ,{\text{ }}dx = 5\sec \theta \tan \theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{\sqrt {{x^2} - 25} }}{x}} dx = \int {\frac{{\sqrt {{{\left( {5\sec \theta } \right)}^2} - 25} }}{{5\sec \theta }}\left( {5\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\frac{{\sqrt {25{{\sec }^2}\theta - 25} }}{{5\sec \theta }}\left( {5\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\frac{{\sqrt {25\left( {{{\sec }^2}\theta - 1} \right)} }}{{5\sec \theta }}\left( {5\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\frac{{\sqrt {25{{\tan }^2}\theta } }}{{5\sec \theta }}\left( {5\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {\frac{{5\tan \theta }}{{5\sec \theta }}\left( {5\sec \theta \tan \theta } \right)} d\theta \cr
& = \int {5{{\tan }^2}\theta } d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = 5\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = 5\tan \theta - 5\theta + C \cr
& {\text{Where }}\sec \theta = \frac{x}{5},{\text{ }}\theta = {\sec ^{ - 1}}\left( {\frac{x}{5}} \right),{\text{ }}\tan \theta = \frac{{\sqrt {{x^2} - 25} }}{5} \cr
& = 5\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right) - 5{\sec ^{ - 1}}\left( {\frac{x}{5}} \right) + C \cr
& = \sqrt {{x^2} - 25} - 5{\sec ^{ - 1}}\left( {\frac{x}{5}} \right) + C \cr} $$
