Answer
$\frac{{\left( {{x^2} + 9} \right)\sqrt {{x^2} + 9} }}{3} - 9\sqrt {{x^2} + 9} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}dx} ,{\text{ }}x = 3\tan \theta \cr
& {\text{Let }}x = 3\tan \theta ,{\text{ }}dx = 3{\sec ^2}\theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}dx} = \int {\frac{{{{\left( {3\tan \theta } \right)}^3}}}{{\sqrt {9 + {{\left( {3\tan \theta } \right)}^2}} }}\left( {3{{\sec }^2}\theta } \right)d\theta } \cr
& = \int {\frac{{{{\left( {3\tan \theta } \right)}^3}}}{{\sqrt {9 + 9{{\tan }^2}\theta } }}\left( {3{{\sec }^2}\theta } \right)d\theta } \cr
& = \int {\frac{{{{\left( {3\tan \theta } \right)}^3}}}{{\sqrt {9\left( {1 + {{\tan }^2}\theta } \right)} }}\left( {3{{\sec }^2}\theta } \right)d\theta } \cr
& {\text{Use the identity }}1 + {\tan ^2}\theta = {\sec ^2}\theta \cr
& = \int {\frac{{{{\left( {3\tan \theta } \right)}^3}}}{{3\sec \theta }}\left( {3{{\sec }^2}\theta } \right)d\theta } \cr
& = 27\int {{{\tan }^3}\theta \sec \theta d\theta } \cr
& = 27\int {{{\tan }^2}\theta \sec \theta \tan \theta d\theta } \cr
& {\text{Use the identity }}1 + {\tan ^2}\theta = {\sec ^2}\theta \cr
& = 27\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta \tan \theta d\theta } \cr
& = 27\int {\overbrace {{{\sec }^2}\theta }^{{u^2}}\overbrace {\left( {\sec \theta \tan \theta d\theta } \right)}^{du}} - 27\int {\sec \theta \tan \theta d\theta } \cr
& {\text{Integrating}} \cr
& = 27\left( {\frac{{{{\sec }^3}\theta }}{3}} \right) - 27\sec \theta + C \cr
& = 9{\sec ^3}\theta - 27\sec \theta + C \cr
& {\text{Write in terms of }}x,{\text{ use the triangle shown below}} \cr
& \sec \theta = \frac{{\sqrt {{x^2} + 9} }}{3},{\text{ then}} \cr
& = 9{\left( {\frac{{\sqrt {{x^2} + 9} }}{3}} \right)^3} - 27\left( {\frac{{\sqrt {{x^2} + 9} }}{3}} \right) + C \cr
& = \frac{{\left( {{x^2} + 9} \right)\sqrt {{x^2} + 9} }}{3} - 9\sqrt {{x^2} + 9} + C \cr} $$
