integral binomial differential

Hiow to evaluate

$\displaystyle \int_0^1{dx \frac{d^{2l-1}}{dx^{2l-1}} \[ (x^2-1)^{2l} \]$

Using the binomial theorem

$\displaystyle \int_0^1{dx \frac{d^{2l-1}}{dx^{2l-1}} \[ \sum_{k=0}^{2l} \frac{(2l)!}{k!(2l-k)!} x^{4l-2k} (-1)^k \]$

l is a non negative integer.

I don't know how to proceed. Is there another way?

Thanks.

Re: integral binomial differential

Move the differential operator into the sum. You'll need to evaluate

$\displaystyle \frac{d^{2l-1}}{dx^{2l-1}} x^{4l-2k}$

which you can easily do by induction. Some terms of the sum may vanish while doing this, I didn't check. Finally, swap the sum and integral. Evaluate the integral. And you should be done.