Answer
$2\sqrt{3}-\frac{10}{3}$
Work Step by Step
\[\begin{align}
& \int_{0}^{1}{\frac{{{x}^{3}}}{\sqrt{3+{{x}^{2}}}}}dx \\
& =\frac{1}{2}\int_{0}^{1}{\frac{{{x}^{2}}}{\sqrt{3+{{x}^{2}}}}\left( 2x \right)}dx \\
& \text{Let }u=3+{{x}^{2}},\text{ }du=2xdx \\
& \text{The new limits of integration are:} \\
& x=1\to u=4 \\
& x=0\to u=3 \\
& \text{Substituting} \\
& \frac{1}{2}\int_{0}^{1}{\frac{{{x}^{2}}}{\sqrt{3+{{x}^{2}}}}\left( 2x \right)}dx=\frac{1}{2}\int_{3}^{4}{\frac{u-3}{\sqrt{u}}du} \\
& =\frac{1}{2}\int_{3}^{4}{\left( {{u}^{1/2}}-3{{u}^{-1/2}} \right)du} \\
& \text{Integrating} \\
& =\frac{1}{2}\left[ \frac{2}{3}{{u}^{3/2}}-6{{u}^{1/2}} \right]_{3}^{4} \\
& =\frac{1}{2}\left[ \frac{2}{3}{{\left( 4 \right)}^{3/2}}-6{{\left( 4 \right)}^{1/2}} \right]-\frac{1}{2}\left[ \frac{2}{3}{{\left( 3 \right)}^{3/2}}-6{{\left( 3 \right)}^{1/2}} \right] \\
& =\frac{1}{2}\left[ \frac{16}{3}-12 \right]-\frac{1}{2}\left[ 2{{\left( 3 \right)}^{1/2}}-6{{\left( 3 \right)}^{1/2}} \right] \\
& =2\sqrt{3}-\frac{10}{3} \\
\end{align}\]
