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Consider the integral $\int_{c[0,3]} \frac{1}{z^2-2z-8}dz$  Notice the function $f(z)=\frac{1}{z-4}$ is holomorphic in $\mathbb{C}/{4}$ which contains $D\overline [-2,3]$. Thus, applying Theorem 4.30:  \begin{align*}  \frac{1}{2\pi i}\int_{c[-2,3]}\frac{\frac{1}{z-4}}{z+2}dz i}\int_{c[-2,3]}\frac{\frac{1}{z-4}}{z+2}dz\\  \end{align*}