Two different solutions to integral

They are both correct: they differ only in terms of their constants of integration.

$$\begin{align} -\frac{1}{2}\ln|-2x| + \color{red}{C} &= -\frac{1}{2}(\ln 2 + \ln |x|) + \color{red}{ C} \\&= -\frac{1}{2}\ln |x| + \color{red}{(C - \frac{1}{2} \ln 2)}\\ &= -\frac{1}{2}\ln |x| + \color{red}{K} \end{align} $$

$C \neq K\;$ but $\;C, K\;$ are constants nonetheless!

TIP: One can always check two apparently different solutions to an integral by differentiating each of them; if the respective derivatives are equal to the original integrand, then you can conclude that the two apparently different solutions are, in fact, solutions that differ only in their constants of integration.

EDIT: You were almost there (in obtaining the adjusted constant of integration), but : $$-\frac{1}{2}\ln{|-2|} + C\; \ne \;-\ln{\frac{1}{\sqrt{2}}} + C$$ Rather, since $\;|-2| \;= \;2,\;$ we have $$-\frac{1}{2}\ln{|-2|}+C \; =\; -\frac{1}{2}\ln 2 + C\;=\; +\ln{\frac{1}{\sqrt{2}}} + C = K$$

The important thing to note is that all of $K$ is a constant term.

They are both correct!

$$ -\frac{1}{2}\ln|-2x| + C = -\frac{1}{2}(\ln 2 + \ln |x|) + C = -\frac{1}{2}\ln |x| + (C - \frac{1}{2} \ln 2) $$

The constant of integration is what "differs" here.

Both of the results is Ok. Note that there is no need that two constants $C$ in the first result and $C$ in the second one are the same. Here, we have $C_1=C_2\times0.5\ln|2|$.