# Calculus: partial fractions

1. Feb 5, 2008

### alexis36

1. The problem statement, all variables and given/known data
Integration of 1/(x^2-5x+6)

2. Relevant equations

3. The attempt at a solution
I know i cannot do ln|x^2-5x+6|
I've tried some form of substitution or intergration by parts, and they don't work.
Should I factor the bottom?

2. Feb 6, 2008

### jhicks

Have you tried partial fractions, as the title suggests? To be brief, you can rewrite $$\frac{1}{x^2-5x+6}$$ as $$\frac{a}{x-2}+\frac{b}{x-3}$$ where you must find a and b.

Last edited: Feb 6, 2008
3. Feb 7, 2008

### alexis36

Okay.. I've solved the question up to a certain point. (I've decided that it's do-able without partial practions..)
however; now I am stuck on an integration:
i might find the integral of
1/(x^2-5x+6)
and now im not sure if i use substituion, integration by parts, or bring the bottom up to the top of the fraction for this integration?

4. Feb 7, 2008

5. Feb 7, 2008

### alexis36

So partial fractions is the way to go I think..
I have
A/(x-3) + B/(x-2) = )A(x-2)+B(x-3))/(x-2)(x-3)
Ive solved for A and B and I got 1 for each of them.
Now I need to go and integrate 1/(x-2) and 1/(x-3) right? and set that equal to (t+c) and then solve the autonomous equation as i would any?

6. Feb 7, 2008

### rocomath

Yes, you would integrate it from here.

7. Feb 7, 2008

### Dick

A=1 and B=1 doesn't work. Think about the signs again.