# Math Help - Rational Fraction decomposition

1. ## 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.

2. 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 $A, B, C$.

3. yes of course! thank you

4. You could also do it without the method of partial fractions:

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

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

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