Integrating Fractions, going insane trying to understand them

Hi, I'm trying to understand how you integrate fractions, I have a couple of examples which I was given, unfortunately the lecturer skipped the actual method and just gave us the answers(Speechless).
By any chance could someone go through the method to solve these?
Thanks in advance

(a) integral ((1-z)/(z² +1))dz

(b) integral ((1)/(z+z²)) dz

(c) integral ((z)/(1-3z²))dz

Re: Integrating Fractions, going insane trying to understand them

Problem 1

Let $I=\int \frac{1-z}{1+z^2}dz$

$=\int \frac{1}{1+z^2}dz - \int \frac{z}{1+z^2}dz$

Substitute $z= \tan \alpha$ and $1+z^2 =\beta$ for the first and second integrals respectively.

$I=\int d\alpha - \frac{1}{2}\int \frac{1}{\beta} \ d\beta = \alpha - \frac{1}{2} \log|\beta|+C$

$= \tan^{-1}(z)- \frac{1}{2}\log|1+z^2|+C$

Problem 2

Let $I=\int \frac{1}{z+z^2}dz = \int \frac{1}{z(1+z)}dz$

$= \int \left( \frac{1}{z}- \frac{1}{z+1}\right)dz = \log|z|-\log|z+1|+C$

$= \log \Big| \frac{z}{z+1}\Big|+C$

Problem 3

Let $I= \int \frac{z}{1-3z^2}dz = \frac{-1}{6}\int \frac{-6z}{1-3z^2}dz$

Substitute $\alpha = 1- 3z^2$ .

$I= \frac{-1}{6}\int \frac{1}{\alpha} d\alpha = -\frac{1}{6}\log|\alpha|+C$

$= \frac{-1}{6}\log|1-3z^2|+C$

Re: Integrating Fractions, going insane trying to understand them

Great thank you!