I want to Evaluate
$$\int_C \frac{3z+1}{z(z-2)^2} dz$$ 
Where C is the above figure-eight contour.
To solve this problem, I split up the given contour $C$ into two simple closed contours $C_1$ and $C_2$. $C_1$ denotes the right-handed closed curve, that has the point $z=2$ inside it, and $C_2$ denotes the left-handed closed curve, that has $z=0$ inside it.
Then, $$\int_C \frac{3z+1}{z(z-2)^2} dz = \int_{C_1} \frac{3z+1}{z(z-2)^2}dz- \int_{C_2} \frac{3z+1}{z(z-2)^2} dz$$
$C_2$ is clockwise or negatively oriented, so that is why I am subtracting the $C_2$ contour integral.
$$\int_{C_{1}} \frac{3z+1}{z(z-2)^2} dz = \int_{C_{1}} \frac{(3z+1)/z}{(z-2)^2} dz$$ I let $f(z) = (3z+1)/z$ which is analytic inside and on $C_1$ since the singularity point $z=0$ is exterior to $C_1$. $z_0=2$ is interior to $C_1$, and thus by Cauchy Integral Formula,
$$\int_{C_1} \frac{3z+1}{z(z-2)^2} dz = \int_{C_1} \frac{(3z+1)/z}{(z-2)^2} dz = 2\pi if'(2) = -\pi i/2$$ since $f'(z) = -1/z^2$.
For $C_2$ integral I also used Cauchy Integral formula except with $f(z) = (3z+1)/(z-2)^2$ which is analytic on and inside $C_2$. Furthermore, $z_0 = 0$ is interior to $C_2$. Thus, $$\int_{C_{2}} \frac{3z+1}{z(z-2)^2} dz = \int_{C_{2}}\frac{(3z+1)/(z-2)^2}{z-0}dz = 2\pi if(0) = \pi i/2$$ since $f(0) = 1/(-2)^2 = 1/4$
Thus,
$$\int_C \frac{3z+1}{z(z-2)^2} dz = -\pi i/2 -\pi i/2 = -\pi i$$
I just want to check if my solution and method are correct for this problem. Thanks!