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