Question

How do you evaluate the integral `int x/(x^2-5x+6)` from 0 to 2?

Answer 1

The definite integral is undefined on the requested range since `ln 0` is undefined.

Explanation:

`x/(x^2-5x+6) = x/((x-3)(x-2))`

`color(white)(x/(x^2-5x+6)) = (3(x-2)-2(x-3))/((x-3)(x-2))`

`color(white)(x/(x^2-5x+6)) = 3/(x-3)-2/(x-2)`

So:

`int_0^2 x/(x^2-5x+6) dx = int_0^2 3/(x-3) - 2/(x-2) dx`

`color(white)(int_0^2 x/(x^2-5x+6) dx) = [color(white)(1/1)3 ln abs(x-3) - 2 ln abs(x-2)color(white)(1/1)]_0^2`

`color(white)(int_0^2 x/(x^2-5x+6) dx) = (3 ln abs(color(blue)(2)-3) - 2 ln abs(color(blue)(2)-2)) - (3 ln abs(color(blue)(0)-3) - 2 ln abs(color(blue)(0)-2))`

`color(white)(int_0^2 x/(x^2-5x+6) dx) = (3 ln 1 - 2 ln 0) - (3 ln 3 - 2 ln 2)`

which is undefined, since `ln 0` is undefined.