Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$. 
I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong. 
Can someone help check my error?

A way to compute this is as follows:
\begin{align*}
\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\mathrm d x &=\int_0^1\frac{4x+x^3-4x}{\sqrt{4+x^2}}\mathrm d x\\
&=\int_0^1x\sqrt{4+x^2}\mathrm d x -2\int_0^1\frac{2x}{\sqrt{4+x^2}}\mathrm d x\\
&=\left.\frac{1}{3}(4+x^2)^{3/2}\right|_0^1-4\left.\left(4+x^2\right)^{1/2}\right|_0^1\\
&=\frac{5\sqrt{5}-8}{3}-4\left(\sqrt{5}-2\right)\\
&=\boxed{\color{blue}{\dfrac{16-7\sqrt{5}}{3}}}
\end{align*}
