Considering the integral $\int \frac{dx}{x\ln |x|}$ I understand the u-substitution way to proceed by substituting $u=\ln |x|; du=dx/x$ and solving. However, I began by trying integration by parts and came up with the following, $u=\frac{1}{\ln |x|}, u’=\frac{-1}{x(\ln |x|)^2}, v’=1/x, v=\ln |x|$ such that $$\int \frac{dx}{x\ln |x|}=\frac{\ln |x|}{\ln |x|}-\int \frac{-1}{x\ln |x|}dx$$ And, if we factor the -1 from the integral on the RHS and then subtract the integral from both sides, haha, presto $0=1$. Help please, I can’t spot my error. Does integration by parts just fail here, and why doesn’t it work out?
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3$\begingroup$ Those are indefinite integrals, and so they differ by a constant of integration. Common mishap when applying integration by parts. $\endgroup$– Simply Beautiful ArtDec 28, 2017 at 1:34
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3$\begingroup$ There's always a $+C$ when you do antidifferentiation. So there's no contradiction. But nor is there any progress in solving the integral. $\endgroup$– Ted ShifrinDec 28, 2017 at 1:34
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$\begingroup$ Ahh, yes there it is! Thank you both @Simply Beautiful Art, and Ted Shifrin $\endgroup$– IsoscelesDec 28, 2017 at 1:38
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$\begingroup$ Also, consider using a better title. Don't be afraid to make the title long, and be sure to include at least some plain words. $\endgroup$– Simply Beautiful ArtDec 28, 2017 at 1:39
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$\begingroup$ @SimplyBeautifulArt I've edited the title with "some plain words". :) $\endgroup$– DeepakDec 28, 2017 at 1:41
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1 Answer
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Try and use the same method for the more fully-formed definate integral $$\int_a^b \frac{dx}{x \log(x)}$$
you can treat $a$ and $b$ as variables if you like, as long as they are independent of $x$. You should understand your error now!
