How do you express #6/(x^2-25)^2# in partial fractions?

1 Answer
Oct 28, 2016

The answer is
#6/(x^2-25)^2=3/(50(x+5)^2)+3/(250(x+5))+3/(50(x-5)^2)-3/(250(x-5))#

Explanation:

we start by factorising the denominator
#(x^2-25)^2=((x+5)(x-5))^2#
So we have#=6(1/((x+5)^2(x-5)^2))#
Then we can do the decomposition in partial fractions
#1/((x+5)^2(x-5)^2)=A/((x+5)^2)+B/(x+5)+C/((x-5)^2)+D/(x-5)#

#=((A(x-5)^2+B(x+5)(x-5)^2+C(x-5)^2+D(x-5)(x+5)^2))/(((x+5)^2(x-5)^2))#

So #1=(A(x-5)^2+B(x+5)(x-5)^2+C(x+5)^2+D(x-5)(x+5)^2))#
Let #x=5# we get #1=100C# #=># #C=1/100#
Let #x=-5# then #1=100A# #=># #A=1/100#
Let #x=0# then #1=25A+125B+25C-125D#
#1=25/100+25/100+125(B-D)#
#B-D=1/250#
Coefficients of #x^3# #=># #0=B+D##=>##B=-D#
#B=1/500# and #D=-1/500#
So the final result is #=6/(100(x+5)^2)+6/(500(x+5))+6/(100(x-5)^2)-6/(500(x-5))#