# How do you use partial fraction decomposition to decompose the fraction to integrate 3/(x^2-6x+8)?

Jun 16, 2015

3int1/(x^2-6x+8)dx = 3int -1/(2(x-2))+1/(2(x-4)

#### Explanation:

$3 \int \frac{1}{{x}^{2} - 6 x + 8} \mathrm{dx}$

You need to factorize the denominator before using partial fraction decomposition

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

$3 \int \frac{1}{\left(x - 2\right) \left(x - 4\right)} \mathrm{dx} = \frac{A}{x - 2} + \frac{B}{x - 4}$

$= \frac{A \left(x - 4\right)}{\left(x - 2\right) \left(x - 4\right)} + \frac{B \left(x - 2\right)}{\left(x - 4\right) \left(x - 2\right)}$

$= A x - 4 A + B x - 2 B = x \left(A + B\right) - 4 A - 2 B$

$A + B = 0$
$- 4 A - 2 B = 1$

So

$A = - \frac{1}{2}$
$B = \frac{1}{2}$

Then

3int1/(x^2-6x+8)dx = 3int -1/(2(x-2))+1/(2(x-4)