Getting different answers for an integral: $\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$ vs $\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$

Question

Problem:

$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$

Using two different methods I am getting two different answers and have trouble finding why.

Method 1:

$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$

$$\int\frac{1}{2}-\frac{3}{2}\left(\frac{1}{x+2}\right)dx$$ $$\int\frac{1}{2}dx-\frac{3}{2}\int\frac{1}{x+2}dx$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{x+2}dx$$ $$x+2=u$$ $$dx=du$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$

$$\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$$

Method 2: $$\int\frac{1}{2}-\frac{3}{2x+4}dx$$

$$\int\frac{1}{2}-3\left(\frac{1}{2x+4}\right)dx$$ $$\int\frac{1}{2}dx-3\int\frac{1}{2x+4}dx$$ $$\frac{1}{2}x-3\int\frac{1}{2x+4}dx$$ $$2x+4=u$$ $$dx=\frac{du}{2}$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$

$$\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$$

Answer 1

Both answers are the same, mind the constant: $$\ln|2x+4|+\color{blue}{C_1}=\ln|2\left(x+2\right)|+\color{blue}{C_1}=\ln|x+2|+ \underbrace{\ln 2 + \color{blue}{C_1}}_{\color{purple}{C_2}} = \ln|x+2|+\color{purple}{C_2}$$

Answer 2

$$\begin{align}\frac12 x + \frac 32 \ln |2x + 4| + C_1 &= \frac12x + \frac32\ln(2\cdot|x+2|) + C_1\\&=\frac12x + \frac32(\ln 2 + \ln|x+2|)+C_1\\&=\frac12x + \frac32 \ln|x+2| + (\frac 32 \ln 2 + C_1)\\&=\frac12x + \frac32\ln|x+2| + C_2\end{align}$$

Answer 3

$\ln|2x+4|=\ln|2(x+2)|=\ln|2|+\ln|x+2|=\ln|x+2|+C$