Find $\int_\Gamma\frac{3z-2}{z^2-z}dz$, where $\Gamma$ encloses point $0$ and $1$.

Question

Find $\int_\Gamma\frac{3z-2}{z^2-z}dz$, where $\Gamma$ is a simple, closed, positively oriented contour enclosing point $0$ and $1$. This question is on Page 197 of Fundamental of Complex Analysis by Snider.

So I understand the solution on the book. The integral equals $6\pi i$.But then I learned Cauchy's integral formula, and I try to do this question in another way, but I got different answers:

Suppose the positively oriented circle around point $1$ is $C_1$, and the one around $0$ is $C_0$.

$\int_\Gamma\frac{3z-z}{z^2-z}dz$=$\int_{C_0}\frac{(3z-z)/z-1)}{z}dz + \int_{C_1}\frac{(3z-z)/z}{z-1}dz$. Since both numerators are analytic on corresponding circles, that integral becomes $2\pi i\times0+2\pi i\times2=4\pi i$. Where did I mess up? Thanks!

Answer 1

$$\oint\limits_{C_0}\frac{\frac{3z-2}{z-1}}zdz+\oint\limits_{C_2}\frac{\frac{3z-2}z}{z-1}dz=2\pi i\left(\left.\frac{3z-2}{z-1}\right|_{z=0}+\left.\frac{3z-2}z\right|_{z=1}\right)=2\pi i\left(2+1\right)=6\pi i$$

Answer 2

I solved the problem using partial fractions which made plugging in trivial. $$ 2\pi if(0) = \oint_C\frac{2}{z}dz = 4\pi i $$ and $$ 2\pi if(1) = \oint_C\frac{1}{z - 1}dz = 2\pi i $$ which gives $f(z) = 6\pi i$