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 :

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

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

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