How do you use partial fraction decomposition to decompose the fraction to integrate #2/(x^3-x^2)#?

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Answer 1 · 2015-08-03T13:19:20

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

Factor the denominator:

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

So we need, #A, B, " and "C# to make:

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

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

#Ax^2-Ax+Bx-B+Cx^2 = 2#

#(A+C)x^2 + (-A+B)x +(-B) = 0x^2+0x+2#

So

#A+C=0#
#-A+B=0#
#-B=2#

And we find:

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

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