Answer
$$3\left( {{x^2} - 2} \right)\sqrt {1 + {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{9{x^3}}}{{\sqrt {1 + {x^2}} }}} dx \cr
& {\text{Using substitution }}x = \tan \theta \cr
& x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{9{x^3}}}{{\sqrt {1 + {x^2}} }}} dx = \int {\frac{{9{{\tan }^3}\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\frac{{9{{\tan }^3}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = \int {9{{\tan }^3}\theta \sec \theta } d\theta \cr
& {\text{Split }}{\tan ^3}\theta \cr
& = \int {9{{\tan }^2}\theta \sec \theta } \tan \theta d\theta \cr
& {\text{Use the pythagorean identity}} \cr
& = 9\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta } \tan \theta d\theta \cr
& = 9\int {{{\sec }^2}\theta \sec \theta \tan \theta } d\theta - 9\int {\sec \theta \tan \theta } d\theta \cr
& {\text{Integrating}} \cr
& = 9\left( {\frac{{{{\sec }^3}\theta }}{3}} \right) - 9\sec \theta + C \cr
& = 3{\sec ^3}\theta - 9\sec \theta + C \cr
& = 3\sec \theta \left( {{{\sec }^2}\theta - 3} \right) + C \cr
& {\text{Where }}\tan \theta = x,{\text{ }}\sec \theta = \sqrt {1 + {x^2}} \cr
& = 3\sqrt {1 + {x^2}} \left( {{{\left( {\sqrt {1 + {x^2}} } \right)}^2} - 3} \right) + C \cr
& = 3\sqrt {1 + {x^2}} \left( {1 + {x^2} - 3} \right) + C \cr
& = 3\left( {{x^2} - 2} \right)\sqrt {1 + {x^2}} + C \cr} $$
