Why does $u$-substitution not work here? $$ \int{\frac{1}{2y}dy} $$
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? 

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