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    • Thread Starter

    away when we did this. I think I understand the concept.

    3x-2/2x+1 is a partial fraction
    3x-2/2x^2-1 is quadratic
    3x-2/(2x-1)^3 is repeated.

    I have a question. x^2+1/x(2x^2+1) : Express in partial fractions.
    How do you go about doing this. Is it quadratic? If so how do you work it out.

    I've got someting saying put it in ax+b/something + c/something

    so do you do:

    ax+b/2x^2+1 + c/x

    I then worked out C to be 1 but how do you get A and B. Am I on the right lines at all ????

    Any help would be appreciated.


    (x^2 + 1) / [x (2x^2 + 1)]
    = (ax + b) / (2x^2 + 1) + c/x.

    Multiply through by x (2x^2 + 1) to get

    (x^2 + 1) = (ax + b)x + c(2x^2 + 1).

    Compare constants: 1 = c.
    Compare coefficients of x: 0 = b.
    Compare coefficients of x^2: 1 = a + 2c, so a = -1.


    (x^2 + 1) / [x (2x^2 + 1)]
    = -x / (2x^2 + 1) + 1/x.
    • Thread Starter


    One more question.


    My final answer came out as

    3/5(x+2) - 3/5(2x+1) - 4/25(x^2+1)

    The first 2 parts were right but the last bit was wrong with the real answer being

    3/5(x+2) - 3/5(2x+1) + (x-1)/5(x^2+1)

    I initially put it in the form

    Ax+b/x^2+1 + c/x+2 + d/2x+1

    I think this is right but I worked out B to be 0 and is this where I went wrong?


    (Original post by Alex H)
    I initially put it in the form

    Ax+b/x^2+1 + c/x+2 + d/2x+1

    I think this is right but I worked out B to be 0 and is this where I went wrong?
    No, ur method were right, but maybe u had some mistakes on calculation.
    Using ur way, I have:
    (ax+b)(x+2)(2x+1) + c(x^2+1)(2x+1) + d(x^2+1)(x+2) = x^3 - 1
    let x = -2, we will have c = 3/5
    let x = -1/2, we have d = -3/5
    let x = 0, we have:
    2b + 3/5 - 2*(3/5) = -1
    2b = -1 + 3/5 = -2/5
    b = -1/5
    let x = 1 we have:
    9(a+b) + 6*(3/5) - 6*(3/5) = 0
    a+b = 0
    a = -b = 1/5
    So u got the answer:
    3/5(x+2) - 3/5(2x+1) + (x-1)/5(x^2 +1)
    Have fun.
    • Thread Starter

    I think it was working out the value for B that I mssed up on. Thanks!!
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Updated: August 18, 2004

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