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

Dec 21, 2016

$\frac{1}{9} \ln | x | - \frac{1}{9} \ln | x - 3 | - \frac{1}{3 \left(x - 3\right)} + C$

#### Explanation:

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

$= \frac{1}{9} \ln | x | + \frac{1}{9} \int \frac{6 - x}{x - 3} ^ 2 \mathrm{dx}$
By substituting $u = x - 3$, $\mathrm{dx} = \mathrm{du}$, or otherwise, the second integral becomes $- \frac{3}{x - 3} - \ln | x - 3 |$ leading to the answer.