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

1 Answer
Roy E.
Dec 21, 2016

#6ln|x|+5/x-6ln|x+1|+C#

Explanation:

Decompose into partial fractions:
#(x-5)/(x^2(x+1))-=(Ax+B)/x^2-6/(x+1)#
giving
#x-5-=(Ax+B)(x+1)-6x^2#
Regrouping:
#0x^2+x^1-5x^0-=(A-6)x^2+(A+B)x^1+Bx^0#
Equating powers of #x#:
#A=6#, #B=-5#
giving the integrand as
#(6x-5)/x^2-6/(x+1)#
#=6/x-5/x^2-6/(x+1)#
The #6# in the partial fractions comes from the cover-up rule, which avoids getting three simultaneous equations (one for each power of #x# in the identity) to solve.

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