Question

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

Answer 1

`3/2ln|x-4| - 3/2ln|x-2| + c`

Explanation:

Begin by factorising the denominator.

`x^2 - 6x + 8 = (x - 2 )(x - 4 )`

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

`rArr 3/((x-2)(x-4)) = A/(x-2) + 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 `rArr B = 3/2`
let x = 2 in (1) : 3 = -2A `rArr A = -3/2`
`rArr3/((x-2)(x-4)) = (3/2)/(x-4) - (3/2)/(x-2)`

and the integral becomes :

`3/2intdx/(x-4) - 3/2intdx/(x-2)`

`= 3/2ln|x-4| - 3/2|x-2| + c`