# How do you use partial fractions to find the integral int (x(2x-9))/(x^3-6x^2+12x-8)dx?

Jan 2, 2017

The answer is =5/(x-2)^2+1/(x-2)+2ln(∣x-2∣)+ C

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(x \ne - 1\right)$

intdx/x=ln∣x∣+C

We start by factorising the denominator

Let $f \left(x\right) = {x}^{3} - 6 {x}^{2} + 12 x - 8$

$f \left(2\right) = 8 - 24 + 24 - 8 = 0$

Therefore,

$\left(x - 2\right)$ is a factor

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 6 {x}^{2} + 12 x - 8$$\textcolor{w h i t e}{a a a a}$∣$x - 2$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣${x}^{2} - 4 x + 4$

$\textcolor{w h i t e}{a a a a a}$$0 - 4 {x}^{2} + 12 x$

$\textcolor{w h i t e}{a a a a a a a}$$- 4 {x}^{2} + 8 x$

$\textcolor{w h i t e}{a a a a a a a a a}$$- 0 + 4 x - 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$+ 4 x - 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$+ 0 - 0$

Therefore,

${x}^{3} - 6 {x}^{2} + 12 x - 8 = \left(x - 2\right) \left({x}^{2} - 4 x + 4\right) = \left(x - 2\right) \left(x - 2\right) \left(x - 2\right)$

so, we can do the decomposition into partial fractions

$\frac{x \left(2 x - 9\right)}{{x}^{3} - 6 {x}^{2} + 12 x - 8} = \frac{x \left(2 x - 9\right)}{x - 2} ^ 3$

$= \frac{A}{x - 2} ^ 3 + \frac{B}{x - 2} ^ 2 + \frac{C}{x - 2}$

$= \frac{A + B \left(x - 2\right) + C {\left(x - 2\right)}^{2}}{x - 2} ^ 3$

$x \left(2 x - 9\right) = A + B \left(x - 2\right) + C {\left(x - 2\right)}^{2}$

Let $x = 2$, $\implies$, $- 10 = A$

Coefficients of ${x}^{2}$, $\implies$, $2 = C$

Let $x = 0$, $\implies$, $0 = A - 2 B + 4 C$

$\implies$, $2 B = A + 4 C = - 10 + 8 = - 2$, $\implies$, $B = - 1$

so,

$\frac{x \left(2 x - 9\right)}{{x}^{3} - 6 {x}^{2} + 12 x - 8} = - \frac{10}{x - 2} ^ 3 - \frac{1}{x - 2} ^ 2 + \frac{2}{x - 2}$

Therefore,

$\int \frac{x \left(2 x - 9\right) \mathrm{dx}}{{x}^{3} - 6 {x}^{2} + 12 x - 8} = - 10 \int \frac{\mathrm{dx}}{x - 2} ^ 3 - \int \frac{\mathrm{dx}}{x - 2} ^ 2 + 2 \int \frac{\mathrm{dx}}{x - 2}$

=5/(x-2)^2+1/(x-2)+2ln(∣x-2∣)+ C