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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 $\int\frac f{f'}$?

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Interesting .. –  user2468 Mar 28 '12 at 16:14

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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.

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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
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@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
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@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

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