Chain rule method doesn't result with same answer as u-sub. Why? $\int \ln \left(2x\right)\,\mathrm{d}x$ is the integral in question. 
I know how to solve it with the chain rule. $\frac{1}{2x}\times 2x = \frac{1}{x}$
But, because I know $u$-sub method, I wish to use it upon this integral. 
$\int \ln \left(2x\right)\,\mathrm{d}x$
$u = 2x$
$du = 2dx \Rightarrow (1/2)du = dx$ 
Thus tranforming the integral into: $(1/2)\int \ln u \, \mathrm{d}u$
Which gives us $\frac{1}{2u}$ and undoing the sub results in $\frac{1}{4x}$ 
Now, this can not be, thus I am at fault and I do not see how.  Please, help explain where I am wrong. 

$$ \frac{1}{2}\int \ln(2x) d2x = \frac{1}{2} \ln(2x)2x - \frac{1}{2}\int2xd\ln(2x) = \frac{\ln(2x)2x}{2} - \int \frac{2x}{x}dx = \frac{\ln(2x)2x}{2} - 2x + C$$
