It is the indefinite integral: $\int \frac{1}{2x-6}$
I am trying to understand it and looking the last step goes from $\frac12 \log(2(x-3))$ to $\frac12 \log(x-3)$
Can someone explain to me why the two just disappears and the answer isn't just $\frac12 \log(2x-6)$
On
Both answers are correct. The difference between the given functions is the constant :) Depends on the method you used. i.e if you take $u=2x-6$ you'll get $\frac12\log(2x-6)$ but if you put $\frac12$ outside of the integral you'll get $\frac12\log(x-3)$.
Both functions have the same derivative
The answer is $\frac{1}{2} \log(2(x-3)) + C = \frac{1}{2} \log(x-3) + \frac{1}{2} \log(2) + C = \frac{1}{2} \log(x-3) + C'$, where $C' = C + \frac{1}{2} \log(2)$ is just another arbitrary constant.
It's more intuitive if you try it with limits. You'll see that the term $\frac{1}{2} \log(2)$ cancels from both ends.