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Math Help - Partial Fractions

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    347

    Question Partial Fractions

    Hi

    How do i solve this partial fraction:

    \frac{4x^2 - 2x}{(x + 1)(x^2 +1)}

    I tried using the same method as I did before...

    \frac{4x^2 - 2x}{(x + 1)(x^2 +1)} = \frac{A}{x + 1} + \frac{B}{x^2 + 1}
    \frac{A(x^2 + 1) + B(x+1)}{(x+1)(x^2+1)}
    4x^2 = 2x = A(x^2 +1) + B(x+1)
    Let x = -1
    4+2 = 2A
    6 = 2A
    A = 3
    4x^2 - 2x = Ax^2 + A + Bx + B
    Differentiate both sides.
    8x - 2 = 2Ax + B
    Substitute in A = 3
    8x - 2 = 6x + B
    2x-2 = B

    I keep on getting 2x - 2 as the value for B even though it's meant to be x - 3.

    Could someone please point out what i did wrong?
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  2. #2
    MHF Contributor
    Joined
    Mar 2007
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    1,240

    Talking

    Shouldn't the second numerator be of the form "Bx + C", due to the fact that the denominator is of degree 2...?

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