Answer
$\sqrt {4{x^2} - 25} - 5{\sec ^{ - 1}}\left( {\frac{{2x}}{5}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {4{x^2} - 25} }}{x}dx} ,{\text{ }}x = \frac{5}{2}\sec \theta \cr
& {\text{Let }}x = \frac{5}{2}\sec \theta ,{\text{ }}dx = \frac{5}{2}\sec \theta \tan \theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{\sqrt {4{x^2} - 25} }}{x}dx} = \int {\frac{{\sqrt {4{{\left( {\frac{5}{2}\sec \theta } \right)}^2} - 25} }}{{\frac{5}{2}\sec \theta }}\left( {\frac{5}{2}\sec \theta \tan \theta } \right)d\theta } \cr
& {\text{Simplify the integrand}} \cr
& = \int {\sqrt {25{{\sec }^2}\theta - 25} \left( {\tan \theta } \right)d\theta } \cr
& = \int {\sqrt {25\left( {{{\sec }^2}\theta - 1} \right)} \left( {\tan \theta } \right)d\theta } \cr
& {\text{Use the identity }}1 + {\tan ^2}\theta = {\sec ^2}\theta \cr
& = \int {\sqrt {25\left( {{{\tan }^2}\theta } \right)} \left( {\tan \theta } \right)d\theta } \cr
& = \int {5\tan \theta \left( {\tan \theta } \right)d\theta } \cr
& = 5\int {{{\tan }^2}\theta d\theta } \cr
& = 5\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = 5\left( {\tan \theta - \theta } \right) + C \cr
& = 5\tan \theta - 5\theta + C \cr
& {\text{Write in terms of }}x,{\text{ use the triangle shown below}} \cr
& \tan \theta = \frac{{\sqrt {4{x^2} - 25} }}{5},{\text{ }}\theta = {\sec ^{ - 1}}\left( {\frac{{2x}}{5}} \right){\text{then}} \cr
& = 5\left( {\frac{{\sqrt {4{x^2} - 25} }}{5}} \right) - 5{\sec ^{ - 1}}\left( {\frac{{2x}}{5}} \right) + C \cr
& = \sqrt {4{x^2} - 25} - 5{\sec ^{ - 1}}\left( {\frac{{2x}}{5}} \right) + C \cr} $$
