# Thread:

1. ## partial fractions

how do I get the answer shown in the attached problem? when I tried it out, I couldn't get the numerator to 1.

Thanks!
Attached Thumbnails

remember that:

$a^2 - b^2 = (a+b)(a-b)$

Can you do it now?

3. Originally Posted by calculus0
how do I get the answer shown in the attached problem? when I tried it out, I couldn't get the numerator to 1.

Thanks!
It will be easier to address your specific trouble if you show all of your working.

4. We write

$\frac{2}{1-x^2}=\frac{2}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}_{.}$

With a bit of algebraic manipulation of the above equality(actually identity), we can find out that

$A = B = 1$.(There are several ways to do this. One of them is writing the right hand side as a single fraction and we get an equation by equating the denominator with 2)

So we have

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