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I could have solved this by substitution, but the ‘n’ is confusing me. How should I proceed?

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closed as off-topic by blub, steven gregory, Paul Frost, José Carlos Santos, Parcly Taxel Aug 13 at 13:22

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    $\begingroup$ If you're integrating with respect to $x$ (where is the $\mbox{d}$x?), then $n$, and also $\ln n$, is just a constant... Can you integrate $\frac{1}{5\ln x}$? $\endgroup$ – StackTD Aug 13 at 12:59
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    $\begingroup$ Probably you misread the question. Make sure to give it another look. The variable wrt which you're supposed to integrate is missing. $\endgroup$ – Paras Khosla Aug 13 at 13:02
  • $\begingroup$ I am sorry, I forgot to add it. The question has been edited, please remove the hold. $\endgroup$ – Aditya Aug 13 at 14:19
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If the question really means $\int\frac{dx}{x\ln n}$ with $n$ a constant with respect to $x$, you get $\frac{\ln |x|}{\ln n}+C=\log_n |x|+C$. If it's a typo for $\int\frac{dx}{x\ln x}$, the substitution $u=\ln x$ gets us to the result $\ln|\ln x|+C$. (Note that in both cases the $C$ is locally constant.)

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    $\begingroup$ what does locally constant mean ? $\endgroup$ – J. W. Tanner Aug 13 at 13:03
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    $\begingroup$ @J.W.Tanner It's allowed to vary across asymptotes. For example, if $y^\prime=1/x$ then $y-\ln|x|$ could be $3$ for $x<0$ but $5$ for $x>0$. $\endgroup$ – J.G. Aug 13 at 13:03
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    $\begingroup$ sincere thanks for the explanation, @J.G. $\endgroup$ – J. W. Tanner Aug 13 at 13:06
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    $\begingroup$ It wasn’t mentioned in the question that n was a constant, but you are probably right, as the answer is correct. Thanks $\endgroup$ – Aditya Aug 13 at 14:22
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If you meant '$x$' instead of '$n$', then substituting $u=\ln x$ gives $\mathrm du=\frac1x\,\mathrm dx$, and thus $$\int\frac{1}{x\ln x}\mathrm dx=\int\frac1u\mathrm du=\ln|u|+C=\ln|\ln(x)|+C.$$

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  • $\begingroup$ No I mean n. I checked the question twice, as I was surprised too. $\endgroup$ – Aditya Aug 13 at 14:20

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