Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is easy to solve integrals of the form $\int\frac{f'}f$ using the defintion of the natural logarithm: $\int \frac{f'(x)}{f(x)}\;\mathrm dx = \ln f(x).\ $ Is there a similar identity for the case $\int\frac f{f'}$?

share|cite|improve this question
Interesting .. – user2468 Mar 28 '12 at 16:14
up vote 8 down vote accepted

Writing $f = e^g$ we have $\int \frac{f}{f'} = \int \frac{1}{g'}$ and this can be a more or less arbitrary integrand so no. Already taking $g = x \ln x - x$ we get a non-elementary integral.

share|cite|improve this answer
I don't understand what argument could be advanced in support of "$1/g'$ can be a more or less arbitrary integrand, so no" that couldn't also be advanced in support of "$g'$ can be a more or less arbitrary integrand, so no". Can you please elaborate? – MJD Mar 28 '12 at 16:19
@MarkDominus: Choose $g$ so that $\displaystyle \frac{1}{g'(x)} = e^{x^2}$. The integral of the latter has no known 'closed form' in terms of elementary functions. – Aryabhata Mar 28 '12 at 16:25
@Aryabhata Your reasoning is too subtle for me to follow. – MJD Mar 28 '12 at 16:42
@MarkDominus: And I don't understand your comment. What exactly do you want clarification on? – Aryabhata Mar 28 '12 at 16:58
@Mark: $g'$ can be a more or less arbitrary integrand, but $\int g' = g$ is no more complicated than $g$; that is, this construction doesn't "increase the complexity." On the other hand, as the above example shows, $\int \frac{1}{g'}$ can be non-elementary even if $g$ is. – Qiaochu Yuan Mar 28 '12 at 17:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.