I am having a problem with this calculus problem:

$\displaystyle \int \frac{1}{y^2-16} dy$

I factored it to: $\displaystyle \int \frac{1}{(y-4)(y+4)} dy$

Then I tried integration by parts: $\displaystyle u=(y-4)^{-1} dv=(y+4)^{-1}$

$\displaystyle du=-(y-4)^{-2} v=ln(y+4)$

$\displaystyle \frac{ln(y+4)}{(y-4)}-\int \frac{ln(y+4)}{(y-4)^2} dy$

So I'm stuck, and I'm just wondering what is the best way to do this?

Peter