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

Question

Elijah Abram · Accepted Answer

#I=3ln(x-3) −9/(x−3) +C# 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#

William Abdalla · Answer

undefined To integrate $ \int \frac{3x}{x^2 - 6x + 9} $ using partial fractions, first factor the denominator $ x^2 - 6x + 9 $ as $ (x - 3)^2 $. Then, express the fraction as $ \frac{A}{x - 3} + \frac{B}{(x - 3)^2} $. Next, find the values of A and B by equating numerators. Finally, integrate each term separately. The integral of $ \frac{A}{x - 3} $ is $ A \ln|x - 3| + C $, and the integral of $ \frac{B}{(x - 3)^2} $ is $ -\frac{B}{x - 3} + C $, where C is the constant of integration.