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

  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. eumyang

    eumyang 1,347
    Homework Helper

    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. 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. Ray Vickson

    Ray Vickson 5,841
    Science Advisor
    Homework Helper

    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. 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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