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\begin{problem}[4.34]  Use the Cauchy Integral Formula(Theorem 4.30) to evaluate the integral in Exercise 4.33 when r=3.   \begin{proof}  Consider the integral $\int_{c[0,3]} $$\int_{c[0,3]}  \frac{1}{z^2-2z-8}dz=\int_{c[0,3]} \frac{dz}{(z+2)(z-4)}$.\\ \frac{dz}{(z+2)(z-4)}$$.  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\\