Confusion with a specific integral: $\displaystyle\int{\dfrac{\ln|2x|}{x\ln|4x|}dx}$

Question

I have this indefinite integral:

$$\int{\dfrac{\ln|2x|}{x\ln|4x|}dx}$$

I solved it this way:

$v=\ln|2x|, \space dv=\dfrac{dx}{x}, \space dx=xdv \\ 2e^v=\ln|4x|$

$ \displaystyle\Rightarrow\int{\frac{xv}{2xe^v}dv}=\frac{1}{2}\int{\frac{v}{e^v}dv}$

$ \displaystyle=\frac{1}{2}\int{ve^{-v}dv} \\ m=v \space, dm=dv \\ dn=e^{-v} \space, n=-e^{-v}$

$ \displaystyle\Rightarrow-\frac{e^{-v}v}{2}+\frac{1}{2}\int{e^{-v}dv}$

$ \displaystyle=-\frac{e^{-v}v+e^{-v}}{2}$

$ \displaystyle=-\frac{e^{-\ln|2x|}(\ln|2x|+1)}{2}$

$ \displaystyle-\frac{\ln|2x|+1}{4x}$

However, different sources show me a big answer with logs of logs, and other times similar answers that are not equivalent. Am I wrong or are they?

Answer 1

$v = \log |2x|$ implies $$\log |4x| = \log |2 (2x)| = \log (2|2x|) = \log 2 + \log |2x| = v + \log 2,$$ using the rule $$\log ab = \log a + \log b, \quad a > 0, \quad b > 0$$ which is clearly satisfied for $a = 2$ and $b = |2x|$ whenever $x \ne 0$.

Consequently your integral after your choice of substitution is $$\int \frac{\log |2x|}{x \log |4x|} \, dx = \int \frac{v}{v + \log 2} \, dv.$$

Answer 2

Let $n=4 x$, then the integral can be transformed into $$ \begin{aligned} I & =\int \frac{\ln \left|\frac{u}{2}\right|}{\frac{u}{4} \ln |u|} \frac{d u}{4} \\ & =\int \frac{\ln |u|-\ln 2}{u \ln |u|} d u \\ & =\int \frac{1}{u} d u-\ln 2 \int \frac{d u}{u \ln |u|} \\ & =\ln |u|-\ln 2 \int \frac{d(\ln |u|)}{\ln |u|} \\ & =\ln |u|-\ln 2 \cdot \ln |\ln | u||+C\\ & =\ln |4x|-\ln 2 \cdot \ln |\ln | 4x||+C \end{aligned} $$