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.

Solution 1:

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.

Solution 2:

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 $$\begin{split} \frac 1 2 \ln(|2y+\frac 1 2|)+C&=\frac 1 2\left( \ln\left|2y+\frac 1 2\right|+\ln 2\right)+C-\frac 1 2 \ln 2\\ &=\frac 1 2\ln\left|2\left(2y+\frac 1 2\right)\right|+C-\frac 1 2 \ln 2\\ &=\frac 1 2\ln\left|4y+1\right|+\underbrace{C-\frac 1 2 \ln 2}_{\text{new constant}}\\ \end{split}$$