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Thread: Quick Question on Partial Fractions Problem

  1. #1
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    Quick Question on Partial Fractions Problem

    Hi
    Can someone tell me if my answer is correct or my book's answers

    1) Resolve into a partial fraction: \frac{x+1}{(x^2+4)(x^2-1)}

    i got \frac{\frac{1}{5}}{(x-1)}-\frac{\frac{1}{5}}{(x^2+4)}

    however book's answers is  \frac{\frac{1}{5}}{(x-1)}-\frac{(x+1)}{5(x^2+4)}<br />
    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone tell me if my answer is correct or my book's answers

    1) Resolve into a partial fraction: \frac{x+1}{(x^2+4)(x^2-1)}

    i got \frac{\frac{1}{5}}{(x-1)}-\frac{\frac{1}{5}}{(x^2+4)}

    however book's answers is  \frac{\frac{1}{5}}{(x-1)}-\frac{(x+1)}{5(x^2+4)}<br />
    P.S
    The book's answer is correct.

    \frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{Ax + B}{(x^2 + 4)} + \frac{Cx + D}{(x^2 - 1)}

     = \frac{(Ax + B)(x^2 - 1) + (Cx + D)(x^2 + 4)}{(x^2 + 4)(x^2 - 1)}.


    So (Ax + B)(x^2 - 1) + (Cx + D)(x^2 + 4) = x + 1

    Ax^3 - Ax + Bx^2 - B + Cx^3 + 4Cx + Dx^2 + 4D = x + 1

    (A + C)x^3 + (B + D)x^2 + (4C - A)x + 4D - B = 0x^3 + 0x^2 + 1x + 1.


    So A + C = 0

    B + D = 0

    4C - A = 1

    4D - B = 1


    Solving these equations simultaneously gives

    A = -\frac{1}{5}, B = -\frac{1}{5}, C = \frac{1}{5}, D = \frac{1}{5}.


    So \frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{-\frac{1}{5}x - \frac{1}{5}}{x^2 + 4} + \frac{\frac{1}{5}x + \frac{1}{5}}{x^2 - 1}

     = -\frac{x + 1}{5(x^2 + 4)} + \frac{x + 1}{5(x^2 - 1)}

     = \frac{x + 1}{5(x - 1)(x + 1)} - \frac{x + 1}{5(x^2 + 4)}

     = \frac{1}{5(x - 1)} - \frac{x + 1}{5(x^2 + 4)}.
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    Quote Originally Posted by Prove It View Post
    The book's answer is correct.

    \frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{Ax + B}{(x^2 + 4)} + \frac{Cx + D}{(x^2 - 1)}
    yeh i made it  = \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 +  4)}.
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    Quote Originally Posted by Paymemoney View Post
    yeh i made it  = \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 +  4)}.
    When dealing with partial fractions, it's always easiest to work with two factors, and then break it down further later if possible.

    Also, your way wouldn't work as it should actually be

    \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 4}.
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