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. 

We usually integrate something like $\int \ln(2x) \, dx$ by parts, as it is quite efficient.
We use the formula $\int u \,dv=uv-\int v\,du$.
Let 
\begin{align*}
u &=\ln(2x) \quad\quad v=x \\
du &=\frac{1}{x}\, dx \,\,\,\,\,\,\,\,\,\, dv=dx
\end{align*}
Then
\begin{align*}
&uv-\int v\,du \\
=&x\ln(2x)-\int 1 \, dx \\
=&x\ln(2x)-x+c.
\end{align*}
If you were to put yourself in a situation where, after a $u$ substitution, all you had to integrate was $\ln(u)$, you would probably integrate that by parts as well (and then back substitute to the same result if done correctly), thus diminishing your efforts to demonstrate an alternate method. 
