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Quick question on partial fractions

  1. May 7, 2008 #1
    I'm a little mixed up on the integration for partial fraction decomposition.

    I basically have x/ x(x^2 + 1)

    I'm wondering for the (x^2 + 1) part, am I to put Ax + B over it because it is a raised power, or since the outside bracket is not squared, it is to only have one variable over it.
     
  2. jcsd
  3. May 7, 2008 #2

    HallsofIvy

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    I thought I had answer this before!

    In general you will need Ax+ B. Notice that you would NOT need that if the denominator were (x+1)^2 so it is NOT a matter of whether "the outside bracket is not squared".

    Of course, for this particular exercise, you should immediately cancel the "x" in numerator and denominator giving 1/(x^2+ 1) which is already in "partial fractions" form so you don't really need the "B".

    But suppose it were (x+2)/(x(x^2+1)). If you were to try (x+2)/(x(x^2+1))= A/x+ B/(x^2+ 1), multiplying on both sides by x(x^2+1) you get x+ 1= A(x^2+ 1)+ Bx= Ax^2+ Bx+ A Equating the corresponding coefficients, A= 0, B= 0, but no way to match the "2".

    Trying (x+2)/(x(x^2+1))= A/x+ (Bx+ C)/(x^2+ 1), multiplying on both sides, you have x+2= A(x^2+1)+ Bx+ C= Ax^2+ Bx+ (A+ C). Now A= 1, B= 0 and A+ C= 2 so A= 1, B= 0, C= 1 works.
     
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