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Partial Fraction Integration problem

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    I have to solve this question and I know that partial fractions is the intended method. I can do the long division easy, which gives:


    Setting up for A and B, I get:


    which produces:

    4x-15= A(x-2)2 + B(x-2)

    From here, I have no idea how to use Gaussian elimination or x values to isolate for A and B. The answers are A=4 and B=-7, but I don't know how they got there. Anyone able to give me a nudge?
  2. jcsd
  3. Jun 22, 2013 #2


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    I don't think your last step is correct.

    If you want to clear the (x-2)^2 term from the denominator, you multiply both sides of the equation by that factor, which leads to the equation,

    4x - 15 = A*(x-2) + B

    which, on expanding, gives

    4x - 15 = A*x - 2*A + B

    equating the coefficients of the various terms,

    4x = A*x

    which implies A = 4, and -15 = B - 2* A or -15 = B - 2*4, or B = -7
  4. Jun 22, 2013 #3
    I guess I didn't understand as well as I thought! Thanks so much for pointing this out SteamKing!
  5. Jun 23, 2013 #4


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    [tex]\frac{4x- 15}{(x- 2)^2}= \frac{A}{(x- 2)}+ \frac{B}{(x- 2)^2}[/tex]
    or, equivalently,
    [tex]4x- 15= A(x- 2)+ B[/tex]
    is to be true for all x, choosing any two values for x will give two equations to solve for A and B.

    Obviously, x= 2 will simplify a lot: 4(2)- 15= -7= B. Taking, say, x= 0 (just because it is simple) gives -15= -2A+ B and we know that B= -7: -15= -2a- 7 so 2a= -7+ 15= 8 and a= 4.

    But any two values of x will give two equations for A and B.
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