# Integrating Fractions, going insane trying to understand them

• Jul 9th 2012, 05:52 AM
MickGrif
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?

(a) integral ((1-z)/(z2 +1))dz

(b) integral ((1)/(z+z2)) dz

(c) integral ((z)/(1-3z2))dz
• Jul 9th 2012, 06:10 AM
sbhatnagar
Re: Integrating Fractions, going insane trying to understand them
Problem 1

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

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

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

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

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

Problem 2

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

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

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

Problem 3

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

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

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

$\displaystyle = \frac{-1}{6}\log|1-3z^2|+C$
• Jul 11th 2012, 03:41 AM
MickGrif
Re: Integrating Fractions, going insane trying to understand them
Great thank you!