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Notice the function $f(z)=\frac{1}{z-4}$ is holomorphic in $\mathbb{C}$$\setminus\left\{4\right\}$ which contains $D\overline [-2,3]$. Thus, we can apply the Cauchy Integral Formula(Theorem 4.30):  \begin{align*}  \frac{1}{2\pi i}\int_{c[-2,3]}\frac{\frac{1}{z-4}}{z+2}dz\\  =2\pi i f(-2)\\  \end{align*}