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Method of Partial Fractions integral help

  1. Nov 26, 2015 #1
    • Moved from a technical forum, so homework template missing
    I have a question where f(x) = 20-2x^2/(x-1)(x+2)^2 and have solved for constants A,B and C.
    A = 2
    B = -4
    C = -4
    I have worked this out myself. Now I am told to compute the indefinite integral and I am getting this answer but apparently it is wrong and I don't understand how?
    My answer: 2ln(abs(x-1))-4ln(x+2)-4ln(abs(x+2)^2)

    Help?
     
  2. jcsd
  3. Nov 26, 2015 #2
    Your constants are right but you're integration isn't correct.

    [tex] \int -\frac{4}{(x+2)^{2}} \neq -4\ln|(x+2)^{2} | [/tex]
     
  4. Nov 26, 2015 #3
    Yea that is the part I am stuck on, I don't know what part of that integration is going wrong.
     
  5. Nov 26, 2015 #4

    Samy_A

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    In general, for ##p\neq-1## and ##a## a constant, ##\int (x+a)^pdx=\frac {1}{p+1} (x+a)^{p+1} + C##
     
  6. Nov 26, 2015 #5

    SammyS

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    Actually it's called the "Method of Partial Fractions".

    To evaluate the integral, ##\displaystyle \int -\,\frac{4}{(x+2)^{2}}dx \, ,\ ## use the substitution u = x+2 .
     
  7. Nov 26, 2015 #6

    Ray Vickson

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    First: you have written
    [tex] f(x) = 20 - \frac{2x^2}{(x-1)}(x+2)^2 [/tex]
    Is that what you meant, or did you want
    [tex] f(x) = \frac{20 - 2 x^2}{(x-1)(x+2)^2}?[/tex]
    If you meant the latter, you need to use parentheses, like this: (20 - 2 x^2)/[(x-1)(x+2)^2].

    Second: what denominators go with the constants A,B,C? We can guess, but we should not need to.

    Third: ##\int (x+2)^{-2} \, dx## does not involve logarithms.
     
  8. Nov 26, 2015 #7

    I like Serena

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    Hi King_Silver! :)

    What's the derivative of ##\frac{1}{x+2}##?
    Does that give us a clue?
     
  9. Nov 28, 2015 #8
    Thanks everyone I actually realised my mistake it was fairly stupid :) fixed it now and got it right!
    it was (4/(x+2))+2ln(abs(x-1))-4ln(abs(x+2))
     
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