Why does $u$-substitution not work here?

Question

$$ \int{\frac{1}{2y}dy} $$

Method 1:

$$\int{\frac{1}{2y}dy} = \frac{1}{2}\int{\frac{1}{y}dy} = \frac{1}{2}\ln|y|+C$$

Method 2 ($u$-substitution):

$$\int{\frac{1}{2y}dy} = \int{\frac{1}{u}dy} = \frac{1}{2}\int{\frac{1}{u}(2)dy}= \frac{1}{2}\int{\frac{1}{u}du}= \frac{1}{2}\ln|u|+C = \frac{1}{2}\ln|2y|+C$$

$$u=2y$$ $$du = 2 dy$$

Why is method 2 wrong?

Answer 1

$$\ln |2y| = \ln 2 + \ln |y|$$

$$ \frac{1}{2}\ln|2y|+C = \frac 12 \ln |y| + \underbrace{\frac 12 \ln 2 + C}_{\large =\, C'}$$ So your results are equivalent up to a constant.