# How do you express 1/(4x^2 - 9) in partial fractions?

##### 1 Answer
Sep 24, 2016

$\frac{1}{4 {x}^{2} - 9} = \frac{\frac{1}{6}}{2 x - 3} + \frac{- \frac{1}{6}}{2 x + 3} .$

#### Explanation:

Let us write $f \left(x\right) = \frac{1}{4 {x}^{2} - 9} = \frac{1}{\left(2 x - 3\right) \left(2 x + 3\right)}$.

So, to convert $f \left(x\right)$ into Partial Fractions, we need constants

A & B in RR, such that,

$f \left(x\right) = \frac{1}{\left(2 x - 3\right) \left(2 x + 3\right)} = \frac{A}{2 x - 3} + \frac{B}{2 x + 3}$.

Now A & B can easily be derived by Heavyside's Cover-up

Method :

$A = {\left[\frac{1}{2 x + 3}\right]}_{2 x - 3 = 0} = \frac{1}{3 + 3} = \frac{1}{6.}$

$B = {\left[\frac{1}{2 x - 3}\right]}_{x = - \frac{3}{2}} = \frac{1}{- 3 - 3} = - \frac{1}{6.}$

Thus,

$\frac{1}{4 {x}^{2} - 9} = \frac{\frac{1}{6}}{2 x - 3} + \frac{- \frac{1}{6}}{2 x + 3} .$