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Integration by Partial Fractions

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    (2x^3-2x+1)/(x^2/3x)
    Find the integral.





    2. The attempt at a solution

    So I've been on this problem for like an hour now and I don't know what I'm doing wrong. So I used long division and got
    2x+ (4x+1)/(x^2-3x)

    ∫2x + ∫(4x+1)/(x^2-3x)
    = x^2 + ∫(4x+1)/(x^2-3x).
    For that next part, I used partial fractions.
    A/x + B/(x-3) = 4x+1
    Solved for a= -1/3
    Solved for b= 13/3

    Put everything together and solving for that integral, I got

    x^2- 1/3ln(x) + 13/3ln(x-3)

    and it's wrong.
    Where am I going wrong? I'm so confused.
    Thank for any help in advance.
     
  2. jcsd
  3. Mar 4, 2014 #2

    vanhees71

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    2016 Award

    Obviously there went something wrong with your partial-fraction calculation. I'd just do the following
    [tex]\frac{2x^3-2x+1}{x^2-3x}=\frac{2x(x^2-3x)+6(x^2-3x)+16x+1}{x^2-3x}=2x+6 + \frac{16x+1}{x^2-3x}.[/tex]
    The first two terms are easily integrated. For the 2nd one you can find the partial fractions with the usual ansatz easily.
     
  4. Mar 4, 2014 #3
    But i was told to use long division first, which was why I started with 2x+ (4x+1)/(x^2-3x).
    And from there, I am confused, as to how I got the answer part wrong.
     
  5. Mar 4, 2014 #4

    vanhees71

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    2016 Award

    What I did is nothing else then long division (written in a way I can achieve with LaTeX).
     
  6. Mar 4, 2014 #5

    HallsofIvy

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    What vanhees71 is trying to tell you is that your long division is wrong. You should have "2x+ 6" as the quotient, not just "2x". And the remainder is "16x+ 1", not "4x+ 1".

    I suspect you did not write the "[itex]0x^2[/itex]" in the dividend and then mistook the "[itex]-6x^2[/itex]" in the division for "[itex]-6x[/itex]".
     
  7. Mar 4, 2014 #6

    Mark44

    Staff: Mentor

    The above has a typo. I'm pretty sure you meant the integrand to be (2x3 - 2x + 1)/(x2 - 3x)
     
  8. Mar 4, 2014 #7
    Ohh, such a simple mistake with the long division. I was forgetting to write the variables I didn't use with 0. Thank you so much everyone.
     
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