Partial fraction: RREF to finish off this problem

**Wolvenmoon** · February 23rd 2011, 01:19 PM

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!

**tonio** · February 23rd 2011, 02:17 PM

Originally Posted by Wolvenmoon

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!

.

**TheCoffeeMachine** · February 24th 2011, 11:28 PM

Originally Posted by Wolvenmoon

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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