Answer
$$\frac{1}{3}\left( {{x^2} - 8} \right)\sqrt {{x^2} + 4} $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx \cr
& {\text{Let }}x = 2\tan \theta ,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx = \int {\frac{{{{\left( {2\tan \theta } \right)}^3}}}{{\sqrt {4 + {{\left( {2\tan \theta } \right)}^2}} }}\left( {2{{\sec }^2}\theta } \right)d\theta } \cr
& = \int {\frac{{8{{\tan }^3}\theta }}{{\sqrt {4 + 4{{\tan }^2}\theta } }}\left( {2{{\sec }^2}\theta } \right)d\theta } \cr
& = \int {\frac{{8{{\tan }^3}\theta }}{{\sqrt {4\left( {1 + {{\tan }^2}\theta } \right)} }}\left( {2{{\sec }^2}\theta } \right)d\theta } \cr
& {\text{Use the pythagorean identity se}}{{\text{c}}^2}\theta = 1 + {\tan ^2}\theta \cr
& = \int {\frac{{8{{\tan }^3}\theta }}{{2\sqrt {{{\sec }^2}\theta } }}\left( {2{{\sec }^2}\theta } \right)d\theta } \cr
& = 8\int {{{\tan }^3}\theta \sec \theta d\theta } \cr
& {\text{Recall that }}{a^{m + n}} = {a^m}{a^n} \cr
& = 8\int {{{\tan }^2}\theta \sec \theta \tan \theta d\theta } \cr
& = 8\int {\left( {{{\sec }^2}\theta + 1} \right)\sec \theta \tan \theta d\theta } \cr
& = 8\int {{{\sec }^2}\theta \sec \theta \tan \theta d\theta } + 8\int {\sec \theta \tan \theta d\theta } \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{8}{3}{\sec ^3}\theta + 8\sec \theta + C \cr
& {\text{We know that }}x = 2\tan \theta ,{\text{ }}\sec \theta = \frac{{\sqrt {{x^2} + 4} }}{2} \cr
& {\text{ = }}\frac{8}{3}{\left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)^3} + 8\left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right) + C \cr
& {\text{ = }}\frac{8}{3}\left( {\frac{{{{\left( {{x^2} + 4} \right)}^{3/2}}}}{8}} \right) + 4\left( {\sqrt {{x^2} + 4} } \right) + C \cr
& = \frac{1}{3}{\left( {{x^2} + 4} \right)^{3/2}} + 4\sqrt {{x^2} + 4} \cr
& {\text{Factoring}} \cr
& {\text{ = }}\frac{1}{3}{\left( {{x^2} + 4} \right)^{1/2}}\left( {{x^2} + 4 - 12} \right) \cr
& {\text{ = }}\frac{1}{3}\left( {{x^2} - 8} \right)\sqrt {{x^2} + 4} \cr} $$
