Answer
$$\frac{{{{\left( {{x^2} + 4} \right)}^{3/2}}}}{3} - \frac{{\sqrt {{x^2} + 4} }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 4} }}} dx \cr
& {\text{substitute }}x = 2\tan \theta ,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr
& = \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 4} }}dx} = \int {\frac{{8{{\tan }^3}\theta }}{{\sqrt {4 + 4{{\tan }^2}\theta } }}} \left( {2{{\sec }^2}\theta d\theta } \right) \cr
& = \int {\frac{{16{{\tan }^3}\theta {{\sec }^2}\theta }}{{2\sqrt {1 + {{\tan }^2}\theta } }}} d\theta \cr
& {\text{pythagorean identity}} \cr
& = \int {\frac{{8{{\tan }^3}\theta {{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} = 8\int {ta{n^3}\theta \sec \theta d\theta } \cr
& = 8\int {\sec \theta \tan \theta {{\tan }^2}\theta d\theta } \cr
& = 8\int {\sec \theta \tan \theta \left( {{{\sec }^2}\theta - 1} \right)d\theta } \cr
& {\text{find the antiderivative}} \cr
& = 8\left( {\frac{{{{\sec }^3}\theta }}{3} - \sec \theta } \right) + C \cr
& {\text{write in terms of }}x \cr
& = \left( {\frac{{8{{\left( {\sqrt {{x^2} + 4} /2} \right)}^3}}}{3} - \frac{{\sqrt {{x^2} + 4} }}{2}} \right) + C \cr
& = \frac{{8{{\left( {{x^2} + 4} \right)}^{3/2}}}}{{24}} - \frac{{\sqrt {{x^2} + 4} }}{2} + C \cr
& = \frac{{{{\left( {{x^2} + 4} \right)}^{3/2}}}}{3} - \frac{{\sqrt {{x^2} + 4} }}{2} + C \cr} $$
