Integrating Fractions, going insane trying to understand them

**MickGrif** · July 9th 2012, 05:52 AM

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

.
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

**sbhatnagar** · July 9th 2012, 06:10 AM

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

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

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

**MickGrif** · July 11th 2012, 03:41 AM

Great thank you!

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