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.

 A: 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
$$\begin{split}
\frac 1 2 \ln(|2y+\frac 1 2|)+C&=\frac 1 2\left( \ln\left|2y+\frac 1 2\right|+\ln 2\right)+C-\frac 1 2 \ln 2\\
&=\frac 1 2\ln\left|2\left(2y+\frac 1 2\right)\right|+C-\frac 1 2 \ln 2\\
&=\frac 1 2\ln\left|4y+1\right|+\underbrace{C-\frac 1 2 \ln 2}_{\text{new constant}}\\
\end{split}$$
A: Use logarithm properties:
\begin{align}
\frac{\ln\, \bigl\lvert 2y+\tfrac12 \bigl\rvert}{2}
&= \frac{\ln\, \Bigl\lvert \bigl( 2y+\tfrac12 \bigr) 
\cdot 2 \cdot \tfrac12 \Bigl\rvert}{2} + C_1 \\
&= \frac{\ln\, \Bigl\lvert \bigl( 4y+1 \bigr) 
\cdot \tfrac12 \Bigl\rvert}{2} + C_1 \\
&= \frac{\ln\, \Bigl( \bigl\lvert 4y+1 \bigr\rvert 
\cdot \bigl\lvert \tfrac12 \bigl\rvert \Bigr)}{2} + C_1 \\
&= \frac{\ln\, \bigl\lvert 4y+1 \bigr\rvert 
+ \ln\, \bigl\lvert \tfrac12 \bigl\rvert}{2} + C_1 \\
&= \frac{\ln\, \bigl\lvert 4y+1 \bigr\rvert}{2} 
+ \frac{\ln\, \bigl\lvert \tfrac12 \bigl\rvert}{2} + C_1 \\
&= \frac{\ln\, \bigl\lvert 4y+1 \bigr\rvert}{2} + C_2,
\end{align}
where
$$
C_2 = \frac{\ln\, \bigl\lvert \tfrac12 \bigl\rvert}{2} + C_1 
= C_1 - \tfrac12 \ln 2
$$
Since the constant of integration is arbitrary for an indefinite integral, either one of these constants could be called $C$. However, it's quite misleading to call them both $C$ since they are explicitly different.
