Tricky partial fraction

1. Aug 26, 2004

Sammon

Would someone be kind enough to show the working for expressing the following in partial fractions:

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

$$\frac{x^2}{(x-1)(x+1)} \equiv 1 + \frac{1}{2(x-1)} - \frac{1}{2(x+1)}$$

Sammon Jnr

2. Aug 26, 2004

Hurkyl

Staff Emeritus
Well, why not add the three fractions in your answer to see if you get the LHS?

3. Aug 26, 2004

humanino

Looks correct. Well, it is correct.

4. Aug 26, 2004

HallsofIvy

Staff Emeritus
I would interpret this as "yes, this is the answer given but I don't know how to arrive at it."

First, actually divide: x2/(x-1)(x+1)= x2/(x2-1)=
1+ 1/(x-1)(x+1).

Now, we are looking for A, B so that 1/(x-1)(x+1)= A/(x-1)+ B/(x+1).

Multiply both sides by (x-1)(x+1) to get 1= A(x+1)+ B(x-1).

Let x= 1: 1= A(2)+ B(0) so A= 1/2.
Let x=-1: 1= A(0)+ B(-2) so B= -1/2.

Putting all of that together, x2/(x-1)(x+1)= 1+ 1/(2(x-1))- 1/(2(x+1)).

5. Aug 26, 2004

humanino

I just noticed you want the working of it ! I would add the RHS as Hurkyl said (but then, you need to know the answer already !), or do

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

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EDIT : once again I am so slow, that I become useless. Better answer from HallsofIvy

6. Aug 27, 2004

Sammon

Many thanks HallsofIvy & humanino!