Answer
$$75\sqrt {{x^2} - 25} + \left( {{x^2} - 25} \right)\sqrt {{x^2} - 25} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^3}}}{{\sqrt {{x^2} - 25} }}dx} \cr
& {\text{substitute }}x = 5\sec \theta ,{\text{ }}dx = 5\sec \theta \tan \theta d\theta \cr
& = \int {\frac{{3{{\left( {5\sec \theta } \right)}^3}}}{{\sqrt {25{{\sec }^2}\theta - 25} }}5\sec \theta \tan \theta d\theta } \cr
& = 3\int {\frac{{125{{\sec }^3}\theta }}{{5\sqrt {{{\sec }^2}\theta - 1} }}5\sec \theta \tan \theta d\theta } \cr
& = 375\int {\frac{{{{\sec }^3}\theta }}{{\sqrt {{{\tan }^2}\theta } }}\sec \theta \tan \theta d\theta } \cr
& {\text{simplify}} \cr
& = 375\int {{{\sec }^4}\theta d\theta } \cr
& {\text{split se}}{{\text{c}}^4}\theta \cr
& = 375\int {{{\sec }^2}\theta {{\sec }^2}\theta d\theta } \cr
& = 375\int {\left( {1 + {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \cr
& t = \tan \theta ,dt = {\sec ^2}\theta d\theta \cr
& = 375\int {\left( {1 + {t^2}} \right)dt} \cr
& = 375t + 125{t^3} + C \cr
& = 375\tan \theta + 125{\tan ^3}\theta + C \cr
& {\text{write in terms of }}x \cr
& = 375\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right) + 125{\left( {\frac{{\sqrt {{x^2} - 25} }}{5}} \right)^3} + C \cr
& = 75\sqrt {{x^2} - 25} + \left( {{x^2} - 25} \right)\sqrt {{x^2} - 25} + C \cr} $$
