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

1 Answer
Mar 9, 2016

#I=3ln(x-3) −9/(x−3) +C#

Explanation:

To integrate #I = 3int (x)/(x^2-6x+9) dx# we will use Partial Fraction Decomposition of the form
#(x)/(x^2-6x+9) = A/(x-3) + B/(x-3)^2#
#(x)/cancel(x^2-6x+9) = (A(x-3) + B)/cancel(x^2-6x+9)# Now determine A and B setting #x=0# and #x=1#
#x= 0: => 0 = -3A + B#; #=> 3A = B#
#x= 1: => 1 = -2A + B# substitute the above and
#1=-2A + 3A; A= 1# and substituting into above #B = 3#
So now our integral becomes:
#I =3int (1/(x-3) + 3/(x-3)^2) dx# Applying linearity
#I = 3(I_1 + I_2) = 3(int 1/(x-3)dx + int 3/(x-3)^2dx)#
#I_1 = int 1/(x-3)dx = ln(x-3)# you can use substitution here
#I_2 = 3int 1/(x-3)^2dx = −3/(x−3)# again use substitution method and apply power rule:
#I = 3(I_1 + I_2)=3(ln(x-3) −3/(x−3))# the final answer need a constant of course:
#I= 3ln(x-3) −9/(x−3) +C#