How does the simplification of the absolute value logarithm work? I am working on an integral and I solved it and got to the red box on the figure.
What I am trying to understand is how they simplified the logarithm function in the red box to the green box.


The trick is that the two $C$ constants are different across the two lines. So this can be confusing.
The second line can be rewritten
$$\begin{split}
\frac 1 2 \ln(|2y+\frac 1 2|)+C&=\frac 1 2\left( \ln\left|2y+\frac 1 2\right|+\ln 2\right)+C-\frac 1 2 \ln 2\\
&=\frac 1 2\ln\left|2\left(2y+\frac 1 2\right)\right|+C-\frac 1 2 \ln 2\\
&=\frac 1 2\ln\left|4y+1\right|+\underbrace{C-\frac 1 2 \ln 2}_{\text{new constant}}\\
\end{split}$$
