Answer
$18 - \frac{{17}}{3}\sqrt {10}$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx \cr
& {\text{Integrating }}\int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx \cr
& {\text{First rewrite the integrand}} \cr
& = \int {\left( {\frac{x}{{\sqrt {9 + {x^2}} }}} \right)} \left( {{x^2}} \right)dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& dv = \frac{x}{{\sqrt {9 + {x^2}} }},{\text{ }}v = \sqrt {9 + {x^2}} \cr
& {\text{Use the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx = {x^2}\left( {\sqrt {9 + {x^2}} } \right) - \int {\left( {\sqrt {9 + {x^2}} } \right)\left( {2x} \right)dx} \cr
& \int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx = {x^2}\sqrt {9 + {x^2}} - \int {\sqrt {9 + {x^2}} \left( {2x} \right)dx} \cr
& \int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx = {x^2}\sqrt {9 + {x^2}} - \left( {\frac{{{{\left( {9 + {x^2}} \right)}^{3/2}}}}{{3/2}}} \right) + C \cr
& \int {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx = {x^2}\sqrt {9 + {x^2}} - \frac{2}{3}{\left( {9 + {x^2}} \right)^{3/2}} + C \cr
& {\text{Therefore}}{\text{,}} \cr
& \int_0^1 {\frac{{{x^3}}}{{\sqrt {9 + {x^2}} }}} dx = \left[ {{x^2}\sqrt {9 + {x^2}} - \frac{2}{3}{{\left( {9 + {x^2}} \right)}^{3/2}}} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& = \left[ {{{\left( 1 \right)}^2}\sqrt {9 + {{\left( 1 \right)}^2}} - \frac{2}{3}{{\left( {9 + {{\left( 1 \right)}^2}} \right)}^{3/2}}} \right] - \left[ {0 - \frac{2}{3}{{\left( {9 + {{\left( 0 \right)}^2}} \right)}^{3/2}}} \right] \cr
& = \left[ {\sqrt {10} - \frac{2}{3}{{\left( {10} \right)}^{3/2}}} \right] - \left[ {0 - \frac{2}{3}\left( {27} \right)} \right] \cr
& = \sqrt {10} - \frac{{20}}{3}\sqrt {10} + 18 \cr
& = 18 - \frac{{17}}{3}\sqrt {10} \cr} $$
