How do you use partial fraction decomposition to decompose the fraction to integrate #(x^2-x+2)/(x(x-1)^2)#?

1 Answer
Jim H
Jul 27, 2015

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

Explanation:

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

Multiply both sides by #x(x-1)^2# (or get a common denominator and compare the numerators -- you'll get to the same place, but one approach may be clearer to you)

#A(x-1)^2+Bx(x-1)+Cx = x^2-x+2#

#Ax^2-2Ax+1+Bx^2-Bx+Cx = x^2-x+2#

#[A+B]x^2 + [-2A-B+C]x +[A] = 1x^2-1x+2#

So we get:

#A+B = 1#
#-2A+B+C=-1#
#A = 2#

Eq 1 and 3 get us #A=2# and #B = -1#

Substituting in Eq 2, we get #C = 2#

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