Integral $\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$ Inspired by some of the greats on this site, I've been trying to improve my residue skills. I've come across the integral
$$\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$$
where $n$ is a positive integer that is at least $2$.
With non-complex methods, I know that the integral is $0$. But I know that it can be done with residue theorem.
The trouble comes in choosing a contour. We're probably going to do some pie-slice contour, perhaps small enough to avoid any of the $2n$th roots of unity, and it's clear that the outer-circle vanishes. But I'm having trouble getting the cancellation for the integral.
Can you help? (Also, do you have a book reference for collections of calculations of integrals with the residue theorem that might have similar examples?)

Just in case someone wonders how it can be done the "normal" way, here it is.
By factorization what we are trying to prove is:
$$
\int_0^\infty \frac{1}{x^n-1} dx = 2 \int_0^\infty \frac{x}{x^{2n}-1} dx
$$
On the right hand side let $x \rightarrow \sqrt{t}$.
