Integral of $1/[(z^{n}+1)(1-z-z^{2})]$ over a circle of large radius

How do you integrate a function in the form $$\frac{1}{(z^{n}+1)(1-z-z^{2})}$$ over a circle of radius $R$?

Also, what happens to this integral as $R \rightarrow \infty$?

Context

This is pulled from this problem set, page 106. It wants me to use the Residue Theorem and them I'm supposed to use Cauchy's integral theorem to create an identity for residues. But I don't know how to find the residue of this function.

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What seems to be giving you problems? There are two general ways you can perform an integral like that. Either you can pick a particular contour, e.g. $\gamma(t)=Re^{2\pi i t}$, or you can use the residue theorem to compute the integral in terms of the residues of the function at the points contained inside the contour. For some problems, you want to do both and then compare the results. The right thing for you to do here will depend on what you know how to do. – Aaron Jan 18 '13 at 0:18
Well this is pulled from the problem set: math.binghamton.edu/sabalka/teaching/09Spring375/Chapter10.pdf (page 106) It wants me to use the Residue Theorem, where Res z=0 = fn. – Anthony Peter Jan 18 '13 at 0:28
@Aaron And them I'm supposed to use Cauchy's integral theorem to create an identity for fn. – Anthony Peter Jan 18 '13 at 0:29
Anthony, where are you having trouble applying the residue theorem? Can you post your work up to that point in detail? – Antonio Vargas Jan 18 '13 at 0:33
@AntonioVargas I just don't know how to find the residue of this function. – Anthony Peter Jan 18 '13 at 0:40

The question you asked is "How do we integrate, and what happens when we take the limit as $R\to \infty$?" which most people would answer with "Use the residue theorem", because that is usually how most contour integrals are most easily done (at least if you want an exact answer and expect that explicit integration will be messy, if not impossible).
However, the question you linked to had the hint of showing that the integral you listed approached zero for large $R$ (and because the only poles of the function are within a circle with small radius, the integral will hence be exactly $0$ for all sufficiently large $R$). This is most easily done by parameterizing a path on the circle, converting the contour integral into a regular integral on, say, $[0,2\pi]$, and then using the fact that $\left|\int_I f dx\right|\leq \int_I \left| f \right| dx$ to bound your integral by something easy to integrate. Because of this, the computation of the hint requires no calculation of residues.
However, the hint implies that the sum of all the residues is equal to zero, and hence (by part 3) $-f_n$ is equal to the sum of the residues not at $z=0$. Because all of the roots of the denominator except the one at $z=0$ are simple, the only truly difficult-to-calculate residue is no longer needed, and the task becomes simplifying the sum of the other residues. This simplification is an exercise in algebra left to the reader.