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Tricky partial fraction

  1. Aug 26, 2004 #1
    Would someone be kind enough to show the working for expressing the following in partial fractions:

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

    I believe the answer is

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

    Thanks in advance,
    Sammon Jnr
     
  2. jcsd
  3. Aug 26, 2004 #2

    Hurkyl

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    Gold Member

    Well, why not add the three fractions in your answer to see if you get the LHS?
     
  4. Aug 26, 2004 #3
    Looks correct. Well, it is correct.
     
  5. Aug 26, 2004 #4

    HallsofIvy

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    I would interpret this as "yes, this is the answer given but I don't know how to arrive at it."

    First, actually divide: x2/(x-1)(x+1)= x2/(x2-1)=
    1+ 1/(x-1)(x+1).

    Now, we are looking for A, B so that 1/(x-1)(x+1)= A/(x-1)+ B/(x+1).

    Multiply both sides by (x-1)(x+1) to get 1= A(x+1)+ B(x-1).

    Let x= 1: 1= A(2)+ B(0) so A= 1/2.
    Let x=-1: 1= A(0)+ B(-2) so B= -1/2.

    Putting all of that together, x2/(x-1)(x+1)= 1+ 1/(2(x-1))- 1/(2(x+1)).
     
  6. Aug 26, 2004 #5
    I just noticed you want the working of it ! I would add the RHS as Hurkyl said (but then, you need to know the answer already !), or do

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

    _______________________________________________________________
    EDIT : once again I am so slow, that I become useless. Better answer from HallsofIvy
     
  7. Aug 27, 2004 #6
    Many thanks HallsofIvy & humanino!

    :smile:
     
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