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?

By the change of variable $x^2+4=t$, we have $2x\,dx=dt$:
$$\int_0^1\frac {x^3}{\sqrt {4+x^2}}\,dx=\frac12\int_4^5\frac{t-4}{\sqrt t}\,dt=\frac13t\sqrt t-4\sqrt t\Big|_4^5=\frac{16-7\sqrt5}3.$$
