# How do you express #(6x^2+1)/(x^2(x-1)^2)# in partial fractions?

##### 1 Answer

Nov 25, 2017

#### Explanation:

#(6x^2+1)/(x^2(x-1)^2) = A/x+B/x^2+C/(x-1)+D/(x-1)^2#

So:

#6x^2+1 = Ax(x-1)^2+B(x-1)^2+Cx^2(x-1)+Dx^2#

Putting

#7 = D#

Putting

#1 = B#

So:

#6x^2+1 = Ax(x-1)^2+(x-1)^2+Cx^2(x-1)+7x^2#

#color(white)(6x^2+1) = Ax(x^2-2x+1)+(x^2-2x+1)+Cx^2(x-1)+7x^2#

#color(white)(6x^2+1) = A(x^3-2x^2+x)+(x^2-2x+1)+C(x^3-x^2)+7x^2#

#color(white)(6x^2+1) = (A+C)x^3+(8-2A-C)x^2+(A-2)x+1#

Hence:

#A=2" "# and#" "C = -2#

So:

#(6x^2+1)/(x^2(x-1)^2) = 2/x+1/x^2-2/(x-1)+7/(x-1)^2#