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?

Hint:
$$\int_0^1\frac{x^3}{\sqrt{4+x^2}}dx=\frac{1}{2}\int_0^1\frac{x^22x}{\sqrt{4+x^2}}dx=\int_0^1\frac{(x^2+4-4)(4+x^2)'}{\sqrt{4+x^2}}dx$$
