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

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Answer 1 · 2016-09-20T19:09:54

#int1/((x+5)^2(x+1))dx=-1/36ln|x+5|+1/(6(x+5))+1/36ln|x-1|+C#

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)#

#=(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))#

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

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

Solving this system of linear equations yields :

#A=-1/36, B=-1/6, C=1/36#

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#

#=-1/36ln|x+5|+1/(6(x+5))+1/36ln|x-1|+C#