Question

How do you integrate `(2x-1)/(x^2-x-6)` using partial fractions?

Answer 1

THe answer is `=ln(x+2)+ln(x-3)+C`

Explanation:

Let's factorise the denominator
`(2x-1)/(x^2-x-6)=(2x-1)/((x+2)(x-3))`
Then we can decompose into partial fractions
Let `(2x-1)/((x+2)(x-3))=A/(x+2)+B/(x-3)`
`=(A(x-3)+B(x+2))/((x+2)(x-3))`
so we can compare the numerators
`2x-1=A(x-3)+B(x+2)`
let `x=3` `=>` `5=5B` `=>` `B=1`
let `x=-2` `=>` `-5=-5A` `=>` `A=1`
and finally we have
`(2x-1)/((x+2)(x-3))=1/(x+2)+1/(x-3)`
Now we can integrate
`int((2x-1)dx)/((x+2)(x-3))=intdx/(x+2)+intdx/(x-3)`
`=ln(x+2)+ln(x-3)+C`