How do you integrate int 1 / ((x+5)^2 (x-1) ) ∫1(x+5)2(x−1) using partial fractions?

1 Answer
Trevor Ryan.
Sep 20, 2016

int1/((x+5)^2(x+1))dx=-1/36ln|x+5|+1/(6(x+5))+1/36ln|x-1|+C∫1(x+5)2(x+1)dx=−136ln|x+5|+16(x+5)+136ln|x−1|+C

Explanation:

Since the denominator has been factorized as a combination of linear factors, one repeated, we may write it in partial fraction form as :

1/((x+5)^2(x+1))=A/(x+5)+B/(x+5)^2+C/(x-1)1(x+5)2(x+1)=Ax+5+B(x+5)2+Cx−1

=(A(x+5)(x-1)+B(x-1)+C(x+5)^2)/((x+5)^2(x-1))=A(x+5)(x−1)+B(x−1)+C(x+5)2(x+5)2(x−1)

=((A+C)x^2+(4A+B+10C)x+(-5A-B+25C))/((x+5)^2(x-1))=(A+C)x2+(4A+B+10C)x+(−5A−B+25C)(x+5)2(x−1)

By comparing terms in the numerator, we now see that :

A+C=0A+C=0
4A+B+10C=04A+B+10C=0
-5A-B+25C=1−5A−B+25C=1

Solving this system of linear equations yields :

A=-1/36, B=-1/6, C=1/36A=−136,B=−16,C=136

Hence the original integral may be written and solved as follows :

int1/((x+5)^2(x+1))dx=-1/36int1/(x+5)dx-1/6int1/(x+5)^2dx+1/36int1/(x-1)dx∫1(x+5)2(x+1)dx=−136∫1x+5dx−16∫1(x+5)2dx+136∫1x−1dx

=-1/36ln|x+5|+1/(6(x+5))+1/36ln|x-1|+C=−136ln|x+5|+16(x+5)+136ln|x−1|+C

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