How do you integrate #2/[x^3-x^2 ]# using partial fractions?

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Answer 1 · 2016-06-10T17:23:39

#int2/(x^3-x^2)dx=int(-2/x-2/x^2+2/(x-1))dx#

#=-2ln|x|+2/x+2ln|x-1|+C#.

Applying the method of partial fractions to rewrite the integral, we may write (denominator factors as linear factors, one repeated) :

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

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

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

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

Hence : #A+C=0 =>A=-C#

#A-B=0 => A=B#

#B=-2#

#therefore A=-2 and C=2#

Hence we may rewrite the integral as :

#int2/(x^3-x^2)dx=int(-2/x-2/x^2+2/(x-1))dx#

#=-2ln|x|+2/x+2ln|x-1|+C#.