Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{ }}\frac{{\sqrt {4 + {x^2}} }}{3}\left( {{x^2} - 8} \right) + C \cr
& \left( {\text{b}} \right){\text{ }}\frac{{\sqrt {4 + {x^2}} }}{3}\left( {{x^2} - 8} \right) + C \cr} $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx \cr
& \left( {\text{a}} \right){\text{ By parts}} \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& {\text{ }}dv = \frac{x}{{\sqrt {4 + {x^2}} }},{\text{ }}v = \frac{1}{2}\int {\frac{{2x}}{{\sqrt {4 + {x^2}} }}dx,{\text{ }}v = \sqrt {4 + {x^2}} } \cr
& \int {udv} = uv - \int {vdu} \cr
& = {x^2}\sqrt {4 + {x^2}} - \int {\sqrt {4 + {x^2}} } \left( {2x} \right)dx \cr
& = {x^2}\sqrt {4 + {x^2}} - \frac{{{{\left( {4 + {x^2}} \right)}^{3/2}}}}{{3/2}} + C \cr
& = {x^2}\sqrt {4 + {x^2}} - \frac{{2{{\left( {4 + {x^2}} \right)}^{3/2}}}}{3} + C \cr
& {\text{Factoring}} \cr
& = \sqrt {4 + {x^2}} \left[ {{x^2} - \frac{2}{3}\left( {4 + {x^2}} \right)} \right] + C \cr
& = \sqrt {4 + {x^2}} \left[ {{x^2} - \frac{8}{3} - \frac{2}{3}{x^2}} \right] + C \cr
& = \sqrt {4 + {x^2}} \left( {\frac{1}{3}{x^2} - \frac{8}{3}} \right) + C \cr
& = \frac{{\sqrt {4 + {x^2}} }}{3}\left( {{x^2} - 8} \right) + C \cr
& \cr
& \left( {\text{b}} \right){\text{ By substitution}} \cr
& {\text{Let }}u = 4 + {x^2},{\text{ }}{x^2} = u - 4,{\text{ }}2xdx = du \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^3}}}{{\sqrt {4 + {x^2}} }}} dx = \int {\frac{{x\left( {u - 4} \right)}}{{\sqrt u }}} \left( {\frac{1}{{2x}}} \right)du \cr
& = \frac{1}{2}\int {\frac{{u - 4}}{{\sqrt u }}} du \cr
& = \frac{1}{2}\int {\left( {{u^{1/2}} - 4{u^{ - 1/2}}} \right)} du \cr
& = \frac{1}{2}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) - 2\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = \frac{1}{3}{u^{3/2}} - 4{u^{1/2}} + C \cr
& = \frac{1}{3}{u^{1/2}}\left( {u - 12} \right) + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{3}\sqrt {4 + {x^2}} \left( {4 + {x^2} - 12} \right) + C \cr
& = \frac{1}{3}\sqrt {4 + {x^2}} \left( {{x^2} - 8} \right) + C \cr} $$
