How do you express $\frac{1}{4 x^{2} - 9}$ $1/(4x^2 - 9)$ in partial fractions?

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Answer 1 · 2016-09-24T05:19:25

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

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

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

$A & B in RR$ , such that,

$f(x)=1/{(2x-3)(2x+3)}=A/(2x-3)+B/(2x+3)$ .

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

Method :

$A=[1/(2x+3)]_(2x-3=0)=1/(3+3)=1/6.$

$B=[1/(2x-3)]_(x=-3/2)=1/(-3-3)=-1/6.$

Thus,

$1/(4x^2-9)=(1/6)/(2x-3)+(-1/6)/(2x+3).$

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