How do you integrate #1/(x^2-4)# using partial fractions?

1 Answer
Narad T.
Oct 28, 2016

The integral is #=1/4(ln(x-2)-ln(x+2))+C#

Explanation:

Let factorise the denominator
#x^2-4=(x-2)(x+2)#
So we have
#1/(x^2-4)=1/((x-2)(x+2))#
So the decomposition into partial fractions is
#1/(x^2-4)=1/((x-2)(x+2))=A/(x-2)+B/(x+2)#

#=(A(x+2)+B(x-2))/((x-2)(x+2)#

So equalising LHS and RHS
#1=A(x+2)+B(x-2)#

If #x=2# then #1=4A##=>##A=1/4#
and #x=-2# then #1=-4B##=># #B=-1/4#

so we have
#1/(x^2-4)=(1/4)/(x-2)+(-1/4)/(x+2)=1/4(1/(x-2)-1/(x+2))#

#int(dx)/(x^2-4)=1/4(int(dx)/(x-2)-int(dx)/(x+2))#

#=1/4(ln(x-2)-ln(x+2))+C#

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