# How do you integrate int 3/ (x^2-6x+8) using partial fractions?

Mar 11, 2016

$\frac{3}{2} \ln | x - 4 | - \frac{3}{2} \ln | x - 2 | + c$

#### Explanation:

Begin by factorising the denominator.

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

Since these factors are linear then the numerators of the partial fractions will be constants , say A and B.

$\Rightarrow \frac{3}{\left(x - 2\right) \left(x - 4\right)} = \frac{A}{x - 2} + \frac{B}{x - 4}$

Multiply through by (x - 2 )(x - 4 )

3 = A(x - 4 ) + B(x - 2 ) .......................................(1)

The aim now is to find the value of A and B. Note that if x = 4 , the term with A will be zero and if x = 2 the term with B will be zero.
let x = 4 in (1) : 3 = 2B $\Rightarrow B = \frac{3}{2}$
let x = 2 in (1) : 3 = -2A $\Rightarrow A = - \frac{3}{2}$
$\Rightarrow \frac{3}{\left(x - 2\right) \left(x - 4\right)} = \frac{\frac{3}{2}}{x - 4} - \frac{\frac{3}{2}}{x - 2}$

and the integral becomes :

$\frac{3}{2} \int \frac{\mathrm{dx}}{x - 4} - \frac{3}{2} \int \frac{\mathrm{dx}}{x - 2}$

$= \frac{3}{2} \ln | x - 4 | - \frac{3}{2} | x - 2 | + c$