Path integral of $1/(z^2-1)$ along $|z|=2$ I would like to calculate the path integral $S$ of $1/(z^2-1)$ along the curve $|z|=2$. 
Using the parametrization $y: [0, 2\pi] \rightarrow \mathbb{C}, y(t)=4e^{it}$ and the partial fraction expansion $1/(z^2-1) = -(1/2)/(z+1) + (1/2) / (z-1)$ we get 
$S=\int_y 1/(z^2-1) dz = \int_y-(1/2)/(z+1) dz+ \int_y(1/2) / (z-1)dz$. 
So to calculate the first integral we would need to determine
$\int_{[0, 2\pi]}e^{it}/(4e^{it}+1)dt$, and a similar expression for the second integral. 
How can we continue from here? 
 A: The function $f(z)=\frac{1}{z^2-1}$ is a function that is $O\left(\frac{1}{z^2}\right)$ as $|z|\to +\infty$, hence the integral of $f(z)$ over $|z|=R$, for large values of $R$, is bounded in absolute value by $\frac{4\pi R}{R^2-1}$, that goes to zero as $R\to +\infty$. On the other hand, by the residue theorem
$$ \oint_{|z|=2}f(z)\,dz = \oint_{|z|=R}f(z)\,dz \tag{1}$$
for any $R>1$, since the enclosed singularities are just the same, $z=\pm 1$. It follows that:
$$ \oint_{|z|=2}f(z)\,dz = \color{red}{0} \tag{2} $$
and the same holds for any meromorphic function with simple poles that is $O\left(\frac{1}{z^2}\right)$ as $|z|\to +\infty$, as soon as the integration contour is a simple plath enclosing every pole.
A: An idea:
$$\frac{\exp(it)}{4\exp(it)+1}=\frac{\exp(it)}{4\exp(it)}\frac{1}{1+\frac{\exp(-it)}{4}}=\frac{1}{4}(\sum_{n\geq 0}(-1)^n\frac{\exp(-int)}{4^n})$$
And now integrate term by term.    
A: As regards your integral
$$\begin{align*}\int_{[0, 2\pi]}\frac{e^{it}}{4e^{it}+1}dt&=\int_{[0, 2\pi]}\frac{e^{it}(4e^{-it}+1)}{|4e^{it}+1|^2}dt\\
&=\int_{[0, 2\pi]}\frac{4+\cos t+i\sin t}{17+8\cos t}dt\\
&=2\int_{[0,\pi]}\frac{4+\cos t}{17+8\cos t}dt\\
&=\frac{1}{4}\left[t+2 \arctan(3/5 \tan(t/2))\right]_0^{\pi}=\frac{\pi}{2}.\end{align*}$$
However, by using Residue Theorem, it is easy to see that the required complex path integral is zero.
A: There's a more elementary approach to this if you really want to do a computation: first do a partial decomposition $\frac{1}{z^2-1}=-\frac{1}{2(z-1)}+\frac{1}{2(z+1)}$. Then notice the contour integral of two terms are exactly $-2\pi i$ and $2\pi i$ by the definition of winding number. Thus, we get the desired result of $0$.
