Rational Fraction decomposition

**Slos** · November 17th 2010, 02:16 AM

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.

**TheCoffeeMachine** · November 17th 2010, 02:31 AM

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 $A, B, C$ .

**Slos** · November 17th 2010, 02:47 AM

yes of course! thank you

**TheCoffeeMachine** · November 17th 2010, 05:45 AM

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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