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Partial Fractions

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Someone teach me how to write equations using the editor here...

    This is part of an integration question

    need partial fractions of 1/(x^2)(x-4)


    2. Relevant equations

    -nil-

    3. The attempt at a solution

    1/(x^2)(x-4)=(Ax+B)/(x^2)+C(x-4)

    Put x=0 therefore B=-1/4
    put x=4 therefore C= 1/16
    Put x=1 therefore A=-1/16

    1/(x^2)(x-4)=(-1/16)(x+4)/x^2 +1/16(x-4)

    I get a different result on Mathematica, someone help
     
  2. jcsd
  3. Jul 14, 2011 #2

    micromass

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    Staff Emeritus
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    Hi Deathfish! :smile:

    If your result is

    [tex]-\frac{x+4}{16x^2}+\frac{1}{16(x+4)}[/tex]

    then it is completely correct!!
    Now, why does mathematica give another result? Well, the above result can be put in a form that is even more convenient. Indeed, the term [itex]\frac{x+4}{16x^2}[/itex] can be "simplified" further as

    [tex]\frac{x+4}{16x^2}=\frac{1}{16x}+\frac{1}{4x^2}[/tex]

    This is a better form since it's more suitable for integration purposes. But your result isn't wrong!
     
  4. Jul 14, 2011 #3

    Mark44

    Staff: Mentor

    The x2 factor in the denominator is not an irreducible quadratic, so there is no need for a term with Ax + b in the numerator.

    I would decompose the expression in this way:
    [tex]\frac{1}{x^2(x - 4)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 4}[/tex]

    How you did it is essentially the same as above, since Ax/x2 is the same as A/x.
     
  5. Oct 25, 2011 #4
    Could the A/X + B/X^2 also be written as Ax+B/X^2. I think you could because splitting it up into Ax/x^2 + B/x^2 is also seen as A/x + B/x^2. Also sorry about dead post reviving...very curious.
     
  6. Oct 26, 2011 #5

    Mark44

    Staff: Mentor

    Yes, but you need parentheses. What you wrote would be considered to be Ax + (B/x2), and I'm certain that's not what you meant.

    Write it as (Ax + B)/x2.
     
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