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

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Answer 1 · 2016-09-24T05:19:25

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

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

So, to convert #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).#