How do you write the partial fraction decomposition of the rational expression # (9x^2 + 1)/(x^2(x − 2)^2)#?

1 Answer
Aug 11, 2018

#(9x^2+1)/(x^2(x-2)^2) = 1/(4x)+1/(4x^2)-1/(4(x-2))+37/(4(x-2)^2)#

Explanation:

Given that we have squared linear factors in the denominator, the decomposition will take a form like:

#(9x^2+1)/(x^2(x-2)^2) = A/x+B/x^2+C/(x-2)+D/(x-2)^2#

Multiplying both sides by #x^2(x-2)^2# this becomes:

#9x^2+1 = Ax(x-2)^2+B(x-2)^2+Cx^2(x-2)+Dx^2#

Putting #x=0# we get:

#1 = 4B" "# so #" "B = 1/4#

Putting #x=2# we get:

#37 = 4D" "# so #" "D=37/4#

Looking at the coefficient of #x#, we have:

#0 = 4A-4B = 4A-1" "# and hence #A=1/4#

Looking at the coefficient of #x^3#, we have:

#0 = A+C" "# and hence #C=-A=-1/4#

So:

#(9x^2+1)/(x^2(x-2)^2) = 1/(4x)+1/(4x^2)-1/(4(x-2))+37/(4(x-2)^2)#