Cauchy Integral Formula for $\oint_{\gamma_i} \frac{z^2+1}{z(z-8)}dz~~~\gamma_i = \mathcal C(3,i), ~~i=1,4,6$ I have a question that i'd like to check my working on. Calculate the integral of $$\oint_{\gamma_i} \frac{z^2+1}{z(z-8)}dz~~~\gamma_i = \mathcal C(3,i), ~~i=1,4,6$$
(a) $\gamma_1$ is the circular contour, positively oriented, with centre 3 and radius 1.
(b) $\gamma_4$ is the circular contour, positively oriented, with centre 3 and radius 4.
(c) $\gamma_6$ is the circular contour, positively oriented, with centre 3 and radius 6.
My attempt:
For a) I wrote $a=0$ and $a=8$ are critical points but are outside the sketch so the answer is $0$
For b) I wrote $a=0$ is inside the circle so I wrote $f(z)=\frac{z^2+1}{z}$ and so = $\int \frac{f(z)}{z-8}$  using the formular we get $2\pi if(8)$ = $\frac{69}{4}\pi i$
and for c) both critical points  $a=0$ and $a=8$ lie in the circle but i'm not sure how to do this bit

For c) you can decompose the fraction 
as follows 
$$\frac{z^2+1}{z(z-8)} =1+- \frac{1}{8z}+\frac{65}{8(z-8)}$$
Thus, 
$$\oint_{\gamma_6} \frac{z^2+1}{z(z-8)}dz=\oint_{\gamma_6} dz-\frac{1}{8}\oint_{\gamma_6}  \frac{dz}{z}+\frac{65}{8}\oint_{\gamma_6} \frac{dz}{(z-8)}\\-\frac{2\pi i}{8}+\frac{2\pi i*65}{8} =\color{red}{16\pi i} $$
Given that $0, 8\in D(3,6)$ and $\oint_{\gamma_6} dz=0$.
