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Math Help - Integration by partial fractions problem

  1. #1
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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.

    Integration by partial fractions problem-calc2.jpg

    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.
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  2. #2
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    Re: Integration by partial fractions problem

    The u in the numerator of your integral should not be there.

    dx=\frac{du}{u}
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    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?
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    Re: Integration by partial fractions problem

    Quote Originally Posted by billb91 View Post
    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.
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    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.
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  6. #6
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    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.
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  7. #7
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    Re: Integration by partial fractions problem

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

    Integration by partial fractions problem-calc2answer.jpg
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