How to find the area where $\frac{1}{z^2-4}$, $z \in \mathbb{C}$ is holomorphic?

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Suppose that you are given a problem of finding the following complex integral:

$$\int_\tau \frac{1}{z^2-4} dz$$

where $\tau = \{z \in \mathbb{C}: |z|=4 \}$. My question is (in the context of this problem): how can we find the area, in which this function - $\frac{1}{z^2-4}$ - is holomorphic?

P.S. I am not interested in finding the integral using Residue Theorem.

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Since: $$\frac{1}{z^2-4}=\frac{1}{4}\left(\frac{1}{z-2}-\frac{1}{z+2}\right)$$ $f(x)$ is a meromorphic function with simple poles in $z=\pm 2$, hence it is a holomorphic function over any region that does not enclose such singularities.

Even if you are not interested, your integral is zero since the residues in the singularities inside $\gamma$ are opposite.