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Homework Help: Integral (Partial-Frac Decomp) SIMPLE?

  1. Oct 26, 2011 #1
    Integral (Partial-Frac Decomp) **SIMPLE?

    1. The problem statement, all variables and given/known data
    [itex]\int^3_2 \frac{-dx}{x^2-1}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    [itex]= \int^3_2 \frac{A}{x} + \frac{B}{x-1} dx[/itex] For some integers A and B.

    [itex] -1 = A(x-1) + B(x+0) [/itex]

    [itex] -1 = Ax - A + Bx [/itex]

    Split into two equations...

    [itex] (1):: -1 = -A [/itex]
    [itex] (2):: A = 1 [/itex]

    [itex] (3):: 0 = A + B [/itex]

    Sub (2) -> (3)

    [itex] (3):: 0 = 1 + B [/itex]
    [itex] (4):: B = -1 [/itex]

    So, from above..

    [itex] \int^3_2 \frac{-dx}{x^2-1} = \int^3_2 \frac{1}{x} - \frac{1}{x-1} dx [/itex]

    [itex] = ln|x| \right|^3_2 - ln|x-1| \right|^3_2 [/itex]

    [itex] = ln(3) - ln(2) - ln(2) + ln(1) [/itex]

    [itex] = ln(3) - 2ln(2) + ln(1) [/itex]

    I entered the final line into the answer box for my course assignment and it said wrong.... So I must be doing something wrong.. =(

    Any help is greatly appreciated. Love you PhysicsForums xx
     
  2. jcsd
  3. Oct 26, 2011 #2
    Re: Integral (Partial-Frac Decomp) **SIMPLE?

    Can you justify this? I don't think it separates like that.
     
    Last edited: Oct 26, 2011
  4. Oct 26, 2011 #3

    LCKurtz

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    Re: Integral (Partial-Frac Decomp) **SIMPLE?

    Wrong at the first step. x2-1 = (x+1)(x-1), not x(x-1).
     
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