Answer
$$\frac{{37\sqrt 3 }}{{72}}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\frac{{{x^3}}}{{{{\left( {3 + {x^2}} \right)}^{5/2}}}}} dx \cr
& {\text{substitute }}x = \sqrt 3 \tan \theta ,{\text{ }}\,\,\,dx = \sqrt 3 {\sec ^2}\theta d\theta \,\,\,\,\theta = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) \cr
& \,\,\,x = 3 \to \theta = {\tan ^{ - 1}}\left( {\frac{3}{{\sqrt 3 }}} \right) = \frac{\pi }{3} \cr
& \,\,\,x = 0 \to \theta = {\tan ^{ - 1}}\left( {\frac{0}{{\sqrt 3 }}} \right) = 0 \cr
& \cr
& {\text{write the integral in terms of }}\theta \cr
& \int_0^3 {\frac{{{x^3}}}{{{{\left( {3 + {x^2}} \right)}^{5/2}}}}} dx = \int_0^{\pi /3} {\frac{{3\sqrt 3 {{\tan }^3}\theta }}{{{{\left( {3 + 3{{\tan }^2}\theta } \right)}^{5/2}}}}\left( {\sqrt 3 {{\sec }^2}\theta } \right)d\theta } \cr
& = \int_0^{\pi /3} {\frac{{3\sqrt 3 {{\tan }^3}\theta }}{{{3^{5/2}}{{\left( {1 + {{\tan }^2}\theta } \right)}^{5/2}}}}\left( {\sqrt 3 {{\sec }^2}\theta } \right)d\theta } \cr
& \cr
& {\text{use the pythagorean identity }}{\tan ^2} + 1 = {\sec ^2}\theta \cr
& = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {\frac{{{{\tan }^3}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{5/2}}}}\left( {{{\sec }^2}\theta } \right)d\theta } \cr
& = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {\frac{{{{\tan }^3}\theta }}{{{{\sec }^5}\theta }}d\theta } \cr
& = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {{{\sin }^3}\theta d\theta } \cr
& = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {{{\sin }^2}\theta \sin \theta d\theta } \cr
& = \frac{{\sqrt 3 }}{3}\int_0^{\pi /3} {\left( {1 - {{\cos }^2}\theta } \right)\sin \theta d\theta } \cr
& {\text{Integrate}} \cr
& = \frac{{\sqrt 3 }}{3}\left[ { - \cos \theta + \frac{{{{\cos }^3}\theta }}{3}} \right]_0^{\pi /3} \cr
& \cr
& {\text{evaluate the limits}} \cr
& = \frac{{\sqrt 3 }}{3}\left[ { - \cos \left( {\frac{\pi }{3}} \right) + \frac{1}{3}{{\cos }^3}\left( {\frac{\pi }{3}} \right)} \right] - \frac{{\sqrt 3 }}{3}\left[ { - \cos \left( 0 \right) + \frac{1}{3}{{\cos }^3}\left( 0 \right)} \right] \cr
& = \frac{{\sqrt 3 }}{3}\left[ { - \frac{1}{2} + \frac{1}{{24}}} \right] - \frac{{\sqrt 3 }}{3}\left[ { - 1 + \frac{1}{3}} \right] \cr
& = - \frac{{11\sqrt 3 }}{{72}} + \frac{{2\sqrt 3 }}{3} \cr
& = \frac{{37\sqrt 3 }}{{72}} \cr} $$
