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

1 Answer
Yonas Yohannes
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

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