# Integrating $\ln(2x)$ by substitution

I've just been looking over some basic calculus and come across the following which I am unable to explain (how the mighty have fallen):

If we integrate $\ln(2x)$ by parts then we quickly get the correct solution $$x\ln(2x) - 2x .$$ However when I try to integrate by substitution I proceed as follows: set $u := 2x \Rightarrow du = 2dx$. Therefore $$\int\ln(2x)dx = \frac 1 2\int\ln(u)du .$$ This is equal to $$\frac{1}{2}(u\ln(u)-u) + c = x\ln(2x)-x + c .$$ Where am I going wrong above?

• $x\ln(2x)-x+c$ is the correct one Nov 4, 2015 at 13:52
• $x\ln(2x)-x + c$ is indeed the right answer. Nov 4, 2015 at 13:52
• You may have performed the integration by parts incorrectly. When I integrate by parts, I get $x\ln 2x - x + C$.
– MPW
Nov 4, 2015 at 13:55

Thank you for the flood of responses in such a short space of time. And now the the explanation of my mistake, and the eternal shame which will follow it. All the above flows from the following basic arithmetic error

$$\frac d{dx}\ln(2x) = 2\cdot\frac{1}{2x}$$ not $$2\cdot\frac{1}{x}$$ as I was doing. Now to change my name, move to cornwall, and lead a quiet life away from Stack Exchange.

• Realizing mistake, posting it as an answer in a humorous fashion, that's more than learning. +1 to keep you hooked on to stack exchange Nov 8, 2015 at 10:35

$$\int \ln(2x)dx$$

By parts:

$\color{gray}{f=\ln(2x)}\text{ and } \color{gray}{df=\frac 1 x dx}$

$\color{gray}{g=x}\text{ and } \color{gray}{dg=dx}$

$$=\boxed{\color{blue}{x\ln(2x)-x+c}}$$

Your first answer "$x\ln(2x) - 2x .$" is $\color{red}{\text{not}}$ correct
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