Because of the duplication of factors we could have
A/x + B/x^2 + C/(x-1) + D/(x-1)^2 = (6x^2 + 1)/(x^2(x-1)^2)
multiply by x^2(x-1)^1
A(x-1)^2 + Bx(x-1)^2 + Cx^2(x-1) + Dx^2 = 6x^2 + 1
let x=0 --> A = 1
let x=1 --> D = 7
then (x-1)^2 + Bx(x-1)^2 + Cx^2(x-1) + 7x^2 = 6x^2 + 1
let x=2 --> 1 + 2B + 4C + 28 = 24+1
or B + 2C = -2 (#1)
let x = -1 --> 4 - 4B - 2C + 7 = 7
or 2B + C = 2 (#2)
2 (#1) - #2
3C = -6
C = -2
then B = 2 form #1
so (6x^2 + 1)/(x^2(x-1)^2)
= 1/x^2 + 2/x - 2/(x-1) + 7/(x-1)^2
Write the partial fraction decomposition of the rational expression.
6x^2+1/x^2(x-1)^2
2 answers
Looks like it should have said
A/x^2 + B/x + C/(x-1) + D/(x-1)^2 = (6x^2 + 1)/(x^2(x-1)^2)
but the solution from there on is correct.
A/x^2 + B/x + C/(x-1) + D/(x-1)^2 = (6x^2 + 1)/(x^2(x-1)^2)
but the solution from there on is correct.