Question

How do you find `int x/(x^2-x-2)dx` using partial fractions?

Answer 1

`int x/(x^2-x-2) dx = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C`

Explanation:

`x/(x^2-x-2) = x/((x-2)(x+1)) = 2/(3(x-2))+1/(3(x+1))`

So:

`int x/(x^2-x-2) dx = int 2/(3(x-2))+1/(3(x+1)) dx`

`color(white)(int x/(x^2-x-2) dx) = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C`