a little more detail...
A bit more about maverick6664's reasoning above: for the case l=l', after you get
\frac{(-1)^l 2 (2l)!}{(l!)^2} \int_0^1 u^l (u-1)^l du,
integrate by parts l more times to get
2 \int_0^1 y^{2l} dy = \frac{2}{2l+1}.
Maybe it's also worth mentioning...