Integral with constant u-substitution

It's the same since you have a constant of integration.. $$ \frac{1}{3} \ln|3x|+c= \frac 13\ln |x|+ \underbrace{\frac{1}{3} \ln|3|+c}_{ \text { is a constant } }=\frac{1}{3} \ln|x|+K$$

Hint: use that $$\ln(3x)=\ln(3)+\ln(x)$$

If you choose the substitution $u=3x\implies du=3\,dx$, then $$\int\frac{dx}{3x}=\int\frac{3\,dx}{3\cdot3x}=\frac13\int\frac{du}{3u}=\frac13\ln|3u|+C=\frac13\ln3+\frac13\ln|u|+C=\color{red}{\frac13\ln|u|+C_1}$$ where $C_1=\frac13\ln3+C$ is another constant.