Answer
$\frac{1}{3}\sqrt {4 + {x^2}} \left( {{x^2} - 8} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx \cr
& \cr
& \left( {\text{a}} \right){\text{Using trigonometric substitution}} \cr
& {\text{Let }}x = 2\tan \theta ,{\text{ }}dx = 2{\sec ^2}\theta d\theta \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
& = \int {\frac{{8{{\tan }^3}\theta }}{{2\sec \theta }}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& = 8\int {{{\tan }^3}\theta } \sec \theta d\theta \cr
& = 8\int {\left( {\frac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}} \right)} \left( {\frac{1}{{\cos \theta }}} \right)d\theta \cr
& = 8\int {\frac{{{{\sin }^3}\theta }}{{{{\cos }^4}\theta }}} d\theta \cr
& = 8\int {\frac{{{{\sin }^2}\theta }}{{{{\cos }^4}\theta }}\left( {\sin \theta } \right)} d\theta \cr
& = 8\int {\frac{{1 - {{\cos }^2}\theta }}{{{{\cos }^4}\theta }}\left( {\sin \theta } \right)} d\theta \cr
& = 8\int {\left( {{{\left( {\cos \theta } \right)}^{ - 4}} - {{\left( {\cos \theta } \right)}^{ - 2}}} \right)\left( {\sin \theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = 8\left[ { - \frac{{{{\cos }^{ - 3}}\theta }}{{ - 3}} + \frac{{{{\left( {\cos \theta } \right)}^{ - 1}}}}{{ - 1}}} \right] + C \cr
& = 8\left[ {\frac{1}{{3{{\cos }^3}\theta }} - \frac{1}{{\cos \theta }}} \right] + C \cr
& = \frac{8}{{3\cos \theta }}\left( {\frac{1}{{{{\cos }^2}\theta }} - 3} \right) + C \cr
& = \frac{8}{3}\sec \theta \left( {{{\sec }^2}\theta - 3} \right) + C \cr
& {\text{We know that }}x = 2\tan \theta ,{\text{ then sec}}\theta = \frac{{\sqrt {4 + {x^2}} }}{2} \cr
& = \frac{8}{3}\left( {\frac{{\sqrt {4 + {x^2}} }}{2}} \right)\left( {\frac{{4 + {x^2}}}{4} - 3} \right) + C \cr
& = \frac{8}{3}\left( {\frac{{\sqrt {4 + {x^2}} }}{2}} \right)\left( {\frac{{4 + {x^2} - 12}}{4}} \right) + C \cr
& = \frac{1}{3}\sqrt {4 + {x^2}} \left( {{x^2} - 8} \right) + C \cr
& \cr
& \left( {\text{b}} \right){\text{Using the substitution }}{u^2} = 4 + {x^2},{\text{ }}2udu = 2xdx \cr
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx = \int {\frac{{{x^2}}}{{\sqrt {4 + {x^2}} }}} \left( x \right)dx = \int {\frac{{\left( {{u^2} - 4} \right)}}{{\sqrt {{u^2}} }}\left( u \right)du} \cr
& = \int {\left( {{u^2} - 4} \right)du} \cr
& = \frac{{{u^3}}}{3} - 4u + C \cr
& {\text{Factoring}} \cr
& = u\left( {\frac{{{u^2}}}{3} - 4} \right) + C \cr
& {\text{Where }}u = \sqrt {4 + {x^2}} \cr
& = \sqrt {4 + {x^2}} \left( {\frac{{{{\left( {\sqrt {4 + {x^2}} } \right)}^2}}}{3} - 4} \right) + C \cr
& = \sqrt {4 + {x^2}} \left( {\frac{{4 + {x^2}}}{3} - 4} \right) + C \cr
& = \frac{{\sqrt {4 + {x^2}} }}{3}\left( {{x^2} - 8} \right) + C \cr
& \cr
& \left( {\text{c}} \right){\text{Using integration by parts}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx,{\text{ }}\int {\frac{x}{{\sqrt {4 + {x^2}} }}\left( {{x^2}} \right)} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}dv = \frac{x}{{\sqrt {4 + {x^2}} }},{\text{ }}v = \sqrt {4 + {x^2}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx = {x^2}\sqrt {4 + {x^2}} - \int {2x\sqrt {4 + {x^2}} } dx \cr
& {\text{Integrating}} \cr
& = {x^2}\sqrt {4 + {x^2}} - \frac{{2{{\left( {4 + {x^2}} \right)}^{3/2}}}}{3} + C \cr
& {\text{Factoring}} \cr
& = \frac{{\sqrt {4 + {x^2}} }}{3}\left[ {3{x^2} - 2\left( {4 + {x^2}} \right)} \right] + C \cr
& = \frac{{\sqrt {4 + {x^2}} }}{3}\left( {{x^2} - 8} \right) + C \cr} $$
