Is $\int \frac{1}{2x} \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$

Note $\frac 12\log|2x| + C_1= \frac 12(\log(2) + \log|x|) + C_1= \frac 12 \log|x| + \frac 12 \log(2) + C_1 = \frac 12\log|x| + C_2$

Notice that $$ \frac{\ln|2x|}{2} = \frac{\ln|2|}{2}+\frac{\ln|x|}{2}$$ and since integration is unique up to constant, resulting integrals are the same. In other words your two results are differing only by constant, thus giving same indefinite integral.

Keep in mind that formally result of indefinite integral is set of functions rather than one specific function. You can think of the integral as a set $$\int f(x) dx = \{F(x); F'(x)=f(x)\}.$$ Then you can quickly see that $$ \frac{\ln|2x|}{2} \in \int \frac{1}{2x} dx $$ $$ \frac{\ln|x|}{2} \in \int \frac{1}{2x} dx $$ since their derivatives are both equal to $\frac{1}{2x}$ (derivative of the constant $\frac{\ln|2|}{2}$ "vanishes"). Even though this notation is rarely used, it gives you better insight into what the indefinite integral really is.