Tricky partial fraction

[Sammon]

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


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


I believe the answer is

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

Thanks in advance,
Sammon Jnr

[Hurkyl]

I believe the answer is

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

[humanino]

Looks correct. Well, it is correct.

[HallsofIvy]

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)).

[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

[Sammon]

Many thanks HallsofIvy & humanino!

:smile: