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Partial Fractions of 1/((3x)(5-x))

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the partial fractions of 1/((3x)(5-x))


    2. Relevant equations



    3. The attempt at a solution
    1/((3x)(5-x))=A/(3x)+B/(5-x)=(A(5-x)+B(3x))/((3x)(5-x))
    1=A(5-x)+B(3x)
    If x=0
    1=A(5-0)+B(3*0)=5A => A=1/5
    If x=5
    1=A(5-5)+B(3*5)=B15 => B=1/15
    So
    1/((3x)(5-x))=1/(15x)+1/(15*(5-x))

    This is what I think it is supposed to be but the answer is slightly different where the third term is negative, 1/((3x)(5-x))=1/(15x)-1/(15*(5-x)). I don't see where the negative sign comes from. It has been a while since I've done partial fractions so I'd like to know if this method is correct.
     
    Last edited: Sep 23, 2011
  2. jcsd
  3. Sep 23, 2011 #2
    I suggest you take out that 3 in the bottom first.

    That is

    [tex]\frac{1}{3}\int \frac{dx}{x(5-x)}[/tex]
     
    Last edited: Sep 23, 2011
  4. Sep 23, 2011 #3
    Pulling the 3 out only seems to make B=1/5 and everything else is the same. Can't see how that makes B negative either
     
  5. Sep 23, 2011 #4
    NO i just suggested it to make multiplication simpler. If you are in doubt, multiply and combine the fractions to see where you went wrong
     
  6. Sep 23, 2011 #5
    Just noticed that the negative came from switching the signs of (5-x) into (x-5) and that I didn't notice the answer had that the other way around too. Problem solved
     
  7. Sep 23, 2011 #6
    Looks to me that you did it correctly. Is there any chance that the answer you were given has x-5 in the denominator rather than 5-x? If not, then the answer given was wrong, it seems.

    So, the other day in another forum, I learned a new method of solving partial fractions by inspection. I've been doing them the traditional way, and someone pointed out the method of residues, which for some reason, I hadn't seen before. But, it hindsight is seems so obvious.

    Anyway, if you don't have degenerate poles, the method is very quick. If you have degenerate poles the method is a little more involved, but not to bad.

    Identify the poles ... in this case x=0 and x=5, the take the limit as x approaches each pole.

    The limit of 1/((3x)(5-x))=A/(3x)+B/(5-x)as x goes to zero is

    1/(5-x)=A=1/5

    And, the limit of 1/((3x)(5-x))=A/(3x)+B/(5-x) as x goes to five is

    1/(3x)=B=1/15

    So, there is a quick way to check, or an even quicker way to cheat :smile:
     
  8. Sep 23, 2011 #7

    SammyS

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    Notice that 1/(5-x) = -1/(x-5) so that [itex]\displaystyle \frac{1}{15x}+\frac{1}{15(5-x)}=\frac{1}{15x}-\frac{1}{15(x-5)}[/itex]
     
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