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

Mar 9, 2016

I=3ln(x-3) −9/(x−3) +C

#### Explanation:

To integrate $I = 3 \int \frac{x}{{x}^{2} - 6 x + 9} \mathrm{dx}$ we will use Partial Fraction Decomposition of the form
$\frac{x}{{x}^{2} - 6 x + 9} = \frac{A}{x - 3} + \frac{B}{x - 3} ^ 2$
$\frac{x}{\cancel{{x}^{2} - 6 x + 9}} = \frac{A \left(x - 3\right) + B}{\cancel{{x}^{2} - 6 x + 9}}$ Now determine A and B setting $x = 0$ and $x = 1$
$x = 0 : \implies 0 = - 3 A + B$; $\implies 3 A = B$
$x = 1 : \implies 1 = - 2 A + B$ substitute the above and
1=-2A + 3A; A= 1 and substituting into above $B = 3$
So now our integral becomes:
$I = 3 \int \left(\frac{1}{x - 3} + \frac{3}{x - 3} ^ 2\right) \mathrm{dx}$ Applying linearity
$I = 3 \left({I}_{1} + {I}_{2}\right) = 3 \left(\int \frac{1}{x - 3} \mathrm{dx} + \int \frac{3}{x - 3} ^ 2 \mathrm{dx}\right)$
${I}_{1} = \int \frac{1}{x - 3} \mathrm{dx} = \ln \left(x - 3\right)$ 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