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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 ..](http://www.wolframalpha.com/input/?i=integration%28f%2Ff%27+dx%29)2012-03-28

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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?2012-03-28
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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.2012-03-28
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    @Aryabhata Your reasoning is too subtle for me to follow.2012-03-28
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    @MarkDominus: And I don't understand your comment. What exactly do you want clarification on?2012-03-28
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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.2012-03-28