Calculus: Evaluate the indefinite integral and simplify your answer - ∫x+1/3x^2+6x-2 dx

I'm having trouble understanding part of this answer. I know the answer is 1/6 ln (3x^2+6x-2)+C but I don't understand where the 1/6 ln comes from. Can anyone explain? Math lab closed before I got off work. Thanks.

∫(x+1)/(3x^2+6x-2) dx , let u = 3x^2+6x-2 ---> du = 6(x+1)dx , so you can integrate

-∫(x + 1)/(3x^2 + 6x - 2) dx

u = 3x^2 + 6x - 2

du/6 = (x + 1) dx

-1/6*∫du/u = -1/6*ln|3x^2 + 6x - 2| + C