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Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRACTION

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Integrate (3x^2-10)/(x^2-4x+4) dx using partial fractions.


    2. Relevant equations

    None

    3. The attempt at a solution

    I tried using A/(x-2) + B/(x-2)^2 but I didnt get a coeffecient of an x^2.

    I've also tried using (Ax+B)/(x-2) + C/(x-2)^2



    Though I honestly thought that the usual A/(x-2) + B/(x-2)^2 would work, How do I know which formula to use. I know you have to do long division if the numerator has an x to a larger value than that of the denominator. I know when you have something like: (x^2-1) or (x^3+2) in the denominator then you use the Ax+B, Cx+D formula.
     
  2. jcsd
  3. Feb 27, 2013 #2

    eumyang

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    Re: Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRAC

    The last thing that you said is correct...
    [tex]\frac{A}{x - 2} + \frac{B}{(x - 2)^2}[/tex]
    ... because the denominator is a linear factor squared. But before you try partial fractions, you have to use long division because the degrees of the numerator and denominator are the same.
     
  4. Feb 27, 2013 #3
    Re: Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRAC

    The degree of the numerator is equal to the degree of the denominator. Try long division before partial fraction decomposition.
     
  5. Feb 27, 2013 #4

    Ray Vickson

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    Re: Indefinite Integral of (3x^2-10)/(x^2-4x+4) dx using PARTIAL FRAC

    Re-write the numerator as
    [tex] 3x^2-10 = 3(x^2-4x+4)+12x - 22.[/tex]
     
  6. Feb 27, 2013 #5
    Ah, so if the degree of the numerator and that of the denominator are the same then you have to use long division, didnt know that. Thanks a lot guys it worked out well.
     
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