# Is there a closed-form expression for $\int (a-b\ln(cx))^{-1} \mathrm{d}x$?

Is there a closed form expression for $$\int\frac{1}{a-b\ln(cx)}\,\mathrm dx\ ?$$

I was wondering how to integrate the above function. I have spent a lot of time on it. First i did an integration by parts until I had to integrate $\ln|a – b \ln(c x)|$ again. Once more, I did an integration by parts and substituted this value back into the original equation. However, now all the terms cancel each other out, effectively giving $0 = 0$. How do I go about integrating $1/(a-b \ln(cx))$? Thanks for the assistance.

Edit: $a$, $b$ and $c$ are constants.

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integrals.wolfram.com suggests that this doesn't have a closed form: it gives an answer in terms of the exponential integral (en.wikipedia.org/wiki/Exponential_integral). – Qiaochu Yuan May 1 '11 at 17:55
Please make the bodies of your posts self-contained. – Arturo Magidin May 1 '11 at 19:04
@Arturo: That may be my fault. The original title was: integrate 1/(a-b*ln(cx)) with respect to x which did not seem descriptive of what the OP appeared to be asking in the question. Sorry if my edit resulted in more confusion rather than less. – cardinal May 1 '11 at 19:07
@cardinal: It wasn't your fault, because the original post did not include the integral he wanted to integrate in the body either, relying instead on information on the subject line. As I've said many times before: think of the subject line like the spine of a book containing its title. What would you think of a book that asks you to go look at the spine to get some basic information without which the text is incomprehensible? – Arturo Magidin May 1 '11 at 19:10

There is no elementary antiderivative. Note that $a - b \ln(c x) = -b \ln(c e^{-a/b} x)$ (assuming $a$, $b$, $c$, $x$ positive), so (after a change of variables) it all reduces to the case $a=0$, $b=c=1$. You can also prove that the integral is non-elementary using the Rothstein-Trager theorem.

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For completeness: the nonelementary integral you'll bump into is $\int\frac{\exp\;u}{u}\mathrm du$, which is the exponential integral $\mathrm{ei}(x)$ that Qiaochu mentions. – J. M. May 1 '11 at 21:05