partial fractions

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June 2nd 2010, 08:41 AM
calculus0

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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

First, factor your denominator.

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!

It will be easier to address your specific trouble if you show all of your working.
June 7th 2010, 07:28 PM
willy0625

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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