# Evaluate $\int \frac{1}{x\ln(n)}dx$ [closed]

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

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

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• 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 $\frac{1}{5\ln x}$? – StackTD Aug 13 at 12:59
• Probably you misread the question. Make sure to give it another look. The variable wrt which you're supposed to integrate is missing. – Paras Khosla Aug 13 at 13:02
• I am sorry, I forgot to add it. The question has been edited, please remove the hold. – Aditya Aug 13 at 14:19

## 2 Answers

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.)

• what does locally constant mean ? – J. W. Tanner Aug 13 at 13:03
• @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 $x>0$. – J.G. Aug 13 at 13:03
• sincere thanks for the explanation, @J.G. – J. W. Tanner Aug 13 at 13:06
• It wasn’t mentioned in the question that n was a constant, but you are probably right, as the answer is correct. Thanks – Aditya Aug 13 at 14:22

If you meant '$$x$$' instead of '$$n$$', then substituting $$u=\ln x$$ gives $$\mathrm du=\frac1x\,\mathrm dx$$, and thus $$\int\frac{1}{x\ln x}\mathrm dx=\int\frac1u\mathrm du=\ln|u|+C=\ln|\ln(x)|+C.$$

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