Integration by partial fractions problem

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February 9th 2013, 10:32 PM
billb91

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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. :)
February 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.

$dx=\frac{du}{u}$
February 9th 2013, 11:50 PM
billb91

Re: Integration by partial fractions problem

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

Not quite. That dx should be du.
February 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.
February 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.
February 10th 2013, 10:10 AM
billb91

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Re: Integration by partial fractions problem

Awesome, thanks for the help! Could anyone verify my answer for this one?

Attachment 26936