Answer
$$\frac{{{{\left( {{x^2} + 4} \right)}^{3/2}}}}{3} - 4\sqrt {{x^2} + 4} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}dx}}{{\sqrt {{x^2} + 4} }}} \cr
& {\text{Use Trigonometric Substitution:}} \cr
& {\text{We set}}{\text{, }}x = 2\tan \theta,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{{x^3}dx}}{{\sqrt {{x^2} + 4} }}} = \int {\frac{{{{\left( {2\tan \theta } \right)}^3}}}{{\sqrt {4 + {{\left( {2\tan \theta } \right)}^2}} }}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{8{{\tan }^3}\theta }}{{\sqrt {4\left( {1 + {{\tan }^2}\theta } \right)} }}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{8{{\tan }^3}\theta }}{{2\sqrt {1 + {{\tan }^2}\theta } }}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = 8\int {\frac{{{{\tan }^3}\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = 8\int {{{\tan }^3}\theta \sec \theta } d\theta \cr
& = 8\int {{{\tan }^2}\theta \sec \theta \tan \theta } d\theta \cr
& = 8\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta \tan \theta } d\theta \cr
& {\text{set }}u = \sec \theta,\,\,\,du = \sec \theta \tan \theta d\theta \cr
& = 8\int {\left( {{u^2} - 1} \right)} du \cr
& {\text{integrating}} \cr
& = 8\left( {\frac{{{u^3}}}{3}} \right) - 8u + C \cr
& {\text{write in terms of }}\theta, \cr
& = \frac{8}{3}{\sec ^3}\theta - 8\sec \theta + C \cr
& {\text{write in terms of }}x,{\text{ sec}}\theta = \frac{{\sqrt {{x^2} + 4} }}{2} \cr
& = \frac{8}{3}{\left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)^3} - 8\left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right) + C \cr
& {\text{simplifying, we get:}} \cr
& = \frac{8}{3}\left[ {\frac{{{{\left( {{x^2} + 4} \right)}^{3/2}}}}{8}} \right] - 4\sqrt {{x^2} + 4} + C \cr
& = \frac{{{{\left( {{x^2} + 4} \right)}^{3/2}}}}{3} - 4\sqrt {{x^2} + 4} + C \cr} $$
