Complex line integral - where the length of the complex number is given I'm trying to study ahead of class so I alleviate some upcoming stress in the semester. I have currently read the chapters on line integrals of complex numbers and am currently trying to deepen the content with some problems given in the workbook.

a)$$\oint_{|z|=1}\frac{dz}{z(z+2)}$$
b) $$\oint_{|z|=3}\frac{dz}{z(z+2)}$$

My general approach on this was to first use partial fraction decomposition to make it a bit easier on the integral.
Meaning that $\frac{1}{z(z+2)}$ becomes $\frac{1}{2z}-\frac{1}{2(z+2)}$ if I didn't make any mistake. The integral would then be $\frac{1}{2}(\ln (2z)-\ln(2(z+2))$. But how do I evaluate that?

I'll consider the integral $\oint_{|z|=1}\frac{dz}{z(z+2)}$. The second one can be done in the similar way (but you need some trick there). I will use Cauchy Integral Theorem (CIT) then.
From CIT we know that if some function $f$ is holomorphic on an open subset $U$ containing a closed disk $D$, then
$$f(a)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(z)}{z-a}dz$$
for any $a$ inside disk $D$. (Note that i didn't formulate the strongest version of this theorem)
Now look at your integrand. If you take $f(z)=\frac{1}{z+2}$, you will see that it is holomorphic in some open neighbourhood of a disk $\{z\in\mathbb{C}:|z|\le1\}$ (for example open disk $\{z\in\mathbb{C}:|z|< \frac{3}{2}\}$).
From CIT you have:
$$f(0)=\frac{1}{2 \pi i} \int_{|z|=1} \frac{f(z)}{z-0}dz=\frac{1}{2 \pi i} \int_{|z|=1} \frac{\frac{1}{z+2}}{z-0}dz=\frac{1}{2 \pi i} \int_{|z|=1} \frac{1}{z(z+2)}dz$$
Thus 
$$\oint_{|z|=1}\frac{dz}{z(z+2)}=2\pi i f(0)=\pi i$$
