How do you integrate #int 1/(x^3-6x^2+9x)# using partial fractions?

1 Answer
Dec 21, 2016

#1/9ln|x|-1/9ln|x-3|-1/(3(x-3))+C#

Explanation:

The integrand may be written #1/(x(x^2-6x+9))#
#-=1/(x(x-3)^2)#
By using the cover-up rule for partial fractions, putting #x=0#, to give the #1/9#:
#-=(1/9)(1/x)+(Ax+B)/(x-3)^2# with #A# and #B# to be found.
Multiplying through by #x(x^2-6x+9)# :
#1-=(1/9)(x-3)^2+Ax^2+Bx# so
#0x^2+0x+1-=(1/9+A)x^2+(-2/3+B)x+1#
For this is identity to be true the coefficients of the powers of #x# must be the same on the two sides. So #A=-1/9#, #B=2/3#.
This gives the integral as
#1/9int1/x+(6-x)/(x-3)^2dx#

#=1/9ln|x|+1/9int(6-x)/(x-3)^2dx#
By substituting #u=x-3#, #dx=du#, or otherwise, the second integral becomes #-3/(x-3)-ln|x-3| # leading to the answer.