Answer
$$\frac{1}{3}{\left( {\sqrt {{x^2} - 25} } \right)^3} + 25\sqrt {{x^2} - 25} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {{x^2} - 25} }}} dx \cr
& {\text{Using substitution }}x = 5\sec \theta \cr
& x = 5\sec \theta ,{\text{ }}dx = 5\sec \theta \tan \theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {{x^2} - 25} }}} dx = \int {\frac{{{{\left( {5\sec \theta } \right)}^3}}}{{\sqrt {{{\left( {5\sec \theta } \right)}^2} - 25} }}} \left( {5\sec \theta \tan \theta } \right)d\theta \cr
& {\text{Simplifying}} \cr
& = \int {\frac{{625{{\sec }^4}\theta \tan \theta }}{{\sqrt {25\left( {{{\sec }^2}\theta - 1} \right)} }}} d\theta \cr
& = \int {\frac{{625{{\sec }^4}\theta \tan \theta }}{{5\sqrt {{{\sec }^2}\theta - 1} }}} d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\frac{{625{{\sec }^4}\theta \tan \theta }}{{5\sqrt {{{\tan }^2}\theta } }}} d\theta \cr
& = 125\int {\frac{{{{\sec }^4}\theta \tan \theta }}{{\tan \theta }}} d\theta \cr
& = 125\int {{{\sec }^4}\theta } d\theta \cr
& = 125\int {{{\sec }^2}\theta {{\sec }^2}\theta } d\theta \cr
& = 125\int {\left( {{{\tan }^2}\theta + 1} \right){{\sec }^2}\theta } d\theta \cr
& = 125\int {{{\tan }^2}\theta {{\sec }^2}\theta } d\theta + 125\int {{{\sec }^2}\theta } d\theta \cr
& {\text{Integrating}} \cr
& = 125\left( {\frac{{{{\tan }^3}\theta }}{3}} \right) + 125\tan \theta + C \cr
& {\text{Where }}\sec \theta = \frac{x}{5},{\text{ }}\tan \theta = \frac{{\sqrt {{x^2} - 25} }}{5} \cr
& = \frac{{125}}{3}\left( {\frac{{{{\left( {\sqrt {{x^2} - 25} } \right)}^3}}}{{125}}} \right) + 125\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right) + C \cr
& = \frac{1}{3}{\left( {\sqrt {{x^2} - 25} } \right)^3} + 25\sqrt {{x^2} - 25} + C \cr} $$
