# Integration by partial fractions problem

• Feb 9th 2013, 10:32 PM
billb91
Integration by partial fractions problem
Hey guys,

I have a homework question that I found a little tricky. I ended up getting an answer but I was hoping for some verification, because I don't feel like I did it correctly. My answer is based on an online integrator, and I know this paper is a bit scattered and difficult to make sense of, so the original equation is on top, the factored version is below the denominator, I did u-substitution to double-check under that, changed it to fractions, and solved for A and B then my answer is on the last line.

Attachment 26930

Also, if it turns out my answer is correct, I'm a tiny bit confused on how the online integrator got such an answer for both separate integrals and a short explanation of how it ended up being that answer would be great. Any help in general is appreciated, I've been stuck on this problem for two days now so it'd be great to knock it out finally. :)
• Feb 9th 2013, 11:20 PM
a tutor
Re: Integration by partial fractions problem
The u in the numerator of your integral should not be there.

$\displaystyle dx=\frac{du}{u}$
• Feb 9th 2013, 11:50 PM
billb91
Re: Integration by partial fractions problem
Would that make the integral I'm left with $\displaystyle \int \frac {1}{u^2-3u+2} \,dx$ or would the denominator be affected by my error also?
• Feb 10th 2013, 12:09 AM
a tutor
Re: Integration by partial fractions problem
Quote:

Originally Posted by billb91
Would that make the integral I'm left with $\displaystyle \int \frac {1}{u^2-3u+2} \,dx$ or would the denominator be affected by my error also?

Not quite. That dx should be du.
• Feb 10th 2013, 12:14 AM
billb91
Re: Integration by partial fractions problem
Oh, sorry, that was what I originally meant. I wasn't being careful when typing it out. :P

Thanks for the help though! I haven't actually gone and solved it yet, but I already know that my new equation will be a LOT easier to solve.
• Feb 10th 2013, 06:45 AM
kspkido
Re: Integration by partial fractions problem
integral(1/u^2-3y+2)du=?
partial fraction first factor denominator : (u-2)(u-1)... then let A/u-2 +B/(u-1)... then multiply [A(u-2) +B(u-1)]/(u-1)(u-2).... first we concentrate on the numerator.. first expand then collect like terms.. you'll end up with Au-2A+Bu-B... note that Au+Bu=0.. this is because if you look at the original equation that 1/u^2-3u+2.... there is no number with the integer 'u'? right? and the -2A-B=1.. since the constant is 1... then do linear of equation.. to get A and B then substitute.. there you have decompose or split the fraction into two.
• Feb 10th 2013, 10:10 AM
billb91
Re: Integration by partial fractions problem
Awesome, thanks for the help! Could anyone verify my answer for this one?

Attachment 26936