Cauchy's integral formula

$\displaystyle \int_{C}\frac{f(z)}{z-z0}dz = 2\pi.i.f(z0) $

Im using cauchy's integral formula, *see above* to solve the following integral

$\displaystyle \int_{C}\frac{z}{z^{2}-1}dz $ where $\displaystyle C $ is a circle of radius 4 centred at 0.

I have deduced that the function $\displaystyle \frac{z}{z^{2}-1} $ is not analytic at $\displaystyle z = +i $ $\displaystyle z = -i $

I let $\displaystyle f(z) = \frac{z}{z+i} $ and $\displaystyle z0 = -i $

so filling into cauchy's integral formula i got

$\displaystyle \int_{C}\frac{z}{(z+i)(z-i)}dz = \frac{f(z)}{z-i} = 2\pi.i.f(i) $

Answer = $\displaystyle \pi.i $

My problem lies in the fact that both $\displaystyle z = +i $ $\displaystyle z = -i $ both lie in $\displaystyle C $. **Therefore do i have to take both singularites into account when evaluating the integral using cauchy's integral formula???**

Thanks,

Piglet