# How does the simplification of the absolute value logarithm work

integrationlogarithms

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.

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.