Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation}
\int \frac{1}{ax + b} dx = \frac{1}{a} \ln{|ax + b|} + C
\end{equation}
the following solution is obtained
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2}\ln{|-2x|} + C
\end{equation}
Why are these solutions different? Which is correct? 
The second solution can be simplified
\begin{equation}
-\frac{1}{2}\ln{|-2x|} + C = -\frac{1}{2}\ln{|-2|} -\frac{1}{2}\ln{|x|} + C= -\ln{\frac{1}{\sqrt{2}}} - \frac{1}{2}\ln{|x|} + C
\end{equation}
but they still differ.

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.
