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Math Help - Partial fraction: RREF to finish off this problem

  1. #1
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    Partial fraction: RREF to finish off this problem

    I have the integral of 1/(x-1)^2(x+1)^2

    This separates to A/(x-1) + B/(x-1)^2 + C/(x+1) + D/(x+1)^2

    Well, going right to where I am with it I have:

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

    The instructor very briefly demonstrated using RREF on problems that got as big as these. Unfortunately I have a TI-89, not an 84. By the time I figured out how to open my matrix editor the right way he was done.

    What do I need to do to tackle this?

    Also, what's the 'other' way?

    Thanks!
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  2. #2
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    Quote Originally Posted by Wolvenmoon View Post
    I have the integral of 1/(x-1)^2(x+1)^2

    This separates to A/(x-1) + B/(x-1)^2 + C/(x+1) + D/(x+1)^2

    Well, going right to where I am with it I have:

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


    This first summand above doesn't look right at all:

    \displaystyle{\frac{1}{(x-1)^2(x+1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}+\frac{D}{(x+1)^2}\Longrighta  rrow}

     1=A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2(x+1)+D(x-1)^2}

    I won't bother solving the system in 4 unknowns which results from comparing

    powers of x above in boths sides and/or choosing some specific values for x, since

    you seem to be working with some calculator method of which I've no idea.

    Tonio



    The instructor very briefly demonstrated using RREF on problems that got as big as these. Unfortunately I have a TI-89, not an 84. By the time I figured out how to open my matrix editor the right way he was done.

    What do I need to do to tackle this?

    Also, what's the 'other' way?

    Thanks!
    .
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  3. #3
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    Quote Originally Posted by Wolvenmoon View Post
    Also, what's the 'other' way?
    You can find the partial fraction by repeatedly putting 1 = \frac{1}{2}\left[(x+1)-(x+1)\right], i.e:


    \displaystyle \begin{aligned} \frac{1}{(x-1)^2(x+1)^2} & = \frac{(x+1)-(x-1)}{2(x-1)^2(x+1)^2} = \frac{1}{2(x-1)^2(x+1)} -\frac{1}{2(x-1)(x+1)^2} \\ & = \frac{(x+1)-(x+1)}{4(x-1)^2(x+1)}-\frac{(x+1)-(x-1)}{4(x-1)(x+1)^2} \\& = \frac{1}{4(x-1)^2}-\frac{1}{4(x-1)(x+1)} -\frac{1}{4(x+1)(x-1)}+\frac{1}{4(x+1)^2} \\& = \frac{1}{4(x-1)^2}-\frac{1}{2(x-1)(x+1)}+\frac{1}{4(x+1)^2} \\& = \frac{1}{4(x-1)^2}-\frac{(x+1)-(x-1)}{4(x-1)(x+1)}+\frac{1}{4(x+1)^2} \\& =  \frac{1}{4(x-1)^2}-\frac{1}{4(x-1)}+\frac{1}{4(x+1)}+\frac{1}{4(x+1)^2}.\end{align  ed}
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