Integration of fractions and cancellation How come $$\int\frac{3}{6+3x}\mathrm{d}x=\mathrm{ln}(2+x)+C$$
and not $\ \mathrm{ln}(6+3x)+C$ ?

I understand you can cancel the fraction but why should this make a difference?

I've tried searching through all my notes on this and online but the terms I am using do not return anything useful.
$$\ln(6+3x)+C=\ln(3[2+x])+C=\ln(2+x)+(\ln 3+C)$$

In other words, both answers are correct, they just use different arbitrary constants.
It is divided by $3$ and thus follows $\log(2+x)$