partial fractions

• June 2nd 2010, 08:41 AM
calculus0
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!
• June 2nd 2010, 09:10 AM
Unknown008

remember that:

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

Can you do it now?
• June 2nd 2010, 07:31 PM
mr fantastic
Quote:

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!

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