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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?

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

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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.

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  • $\begingroup$ 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 $\endgroup$
    – Shailesh
    Nov 8, 2015 at 10:35
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$$\int \ln(2x)dx$$

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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