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Math Help - Rational Fraction decomposition

  1. #1
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    Rational Fraction decomposition

    Hi there,

    It is probably a really easy question but i am a bit rusty.

    I would like basically to decompose :

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

    I thus need to breakdown into simple elements:

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

    Now I can't remember how to find out A, B and C.
    I've already checked online and I understood that, for example to find B:
    I should multiply the first fraction (1) by x+1 and then change x for -1. but if i do that it will cancel everything.
    on the other hand, I can't understand how to find A.

    Pretty sure it is quiet basic question but does anyone could explain me the method for this fraction.

    Thanks in advance.
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  2. #2
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    Take the LHS over a common denominator:

    \displaystyle \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1} = \frac{(x-1)(x+1)A+x(x-1)B+x(x+1)C}{x(x+1)(x-1)}

    So we have:

    \displaystyle \frac { 1 }{ x(x -1)(x+1) } \equiv \frac{(x-1)(x+1)A+x(x-1)B+x(x+1)C}{x(x+1)(x-1)}

    Therefore:

    \displaystyle 1 \equiv (x-1)(x+1)A+x(x-1)B+x(x+1)C

    Put x = 0, -1, 1, to respectively find A, B, C.
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  3. #3
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    yes of course! thank you
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  4. #4
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    You could also do it without the method of partial fractions:

    \displaystyle \frac {1}{ x(x -1)(x+1) } = \frac{(x+1)-(x-1)}{2x(x-1)(x+1)} = \frac{1}{2x(x-1)}-\frac{1}{2x(x+1)} = \frac{x-(x-1)}{2x(x-1)}-\frac{(x+1)-x}{2x(x+1)}

    \displaystyle  \frac{x}{2x(x-1)}-\frac{x-1}{2x(x-1)}-\frac{x+1}{2x(x+1)}+\frac{x}{2x(x+1)} \displaystyle =  \frac{1}{2(x-1)}-\frac{1}{2x}-\frac{1}{2x}+\frac{1}{2(x+1)}

    \displaystyle \therefore ~ \frac{1}{x(x -1)(x+1)} = \frac{1}{2(x-1)}-\frac{1}{2x}-\frac{1}{2x}+\frac{1}{2(x+1)} = \frac{1}{2(x-1)}-\frac{1}{x}+\frac{1}{2(x+1)}.
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