How do you express #2/(x^3-x^2)# in partial fractions?

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Answer 1 · 2016-03-23T00:13:46

#2/(x^3-x^2) = 2/(x-1)-2/x-2/x^2#

The denominator splits into factors as: #x^3-x^2 = x^2(x-1)#

So we solve for a partial fraction decomposition of the form:

#2/(x^3-x^2)#

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

#=(Ax^2+Bx(x-1)+C(x-1))/(x^3-x^2)#

#=((A+B)x^2+(C-B)x-C)/(x^3-x^2)#

Equating coefficients, we get:

#{ (A+B = 0), (C-B = 0), (-C=2) :}#

Hence we find:

#{ (C = -2), (B = -2), (A = 2) :}#

So:

#2/(x^3-x^2) = 2/(x-1)-2/x-2/x^2#