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Is there any way to find $$\int \frac1{f(x)}\mathrm dx$$ in terms of $\int f(x) \mathrm dx$, $f(x)$ and its derivatives?

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3 Answers 3

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In general, no. $\int x\exp(-x)\mathrm dx$ is elementary, while $\int \frac{\exp(x)}{x}\mathrm dx$ is not expressible in terms of elementary functions.

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    $\begingroup$ For a somewhat simpler example along the same lines, take $f = \log$. $\endgroup$
    – lhf
    Sep 6, 2011 at 16:16
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    $\begingroup$ @lhf: Simpler in form only, since a change variables gives $$ \int\frac{\mathrm{d}x}{\log(x)}=\int\frac{\exp(u)\;\mathrm{d}u}{u} $$ $\endgroup$
    – robjohn
    Sep 6, 2011 at 16:30
  • $\begingroup$ @robjohn: hence "along the same lines" methinks. ;) $\endgroup$ Sep 6, 2011 at 16:32
  • $\begingroup$ @J. M.: fair enough. $\endgroup$
    – robjohn
    Sep 6, 2011 at 16:35
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How about: $\int x \,dx$ is a polynomial, a rational function, an algebraic function, while $\int dx/x$ is none of these.

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As others have stated, the answer is no. However, assuming $f$ is analytic and $f(x_0) \ne 0$, you do have a Taylor series for an antiderivative of $1/f(x)$ in a neighbourhood of $x_0$, which can be expressed in terms of the values of $f$ and its derivatives at $x_0$:

$$ \matrix{ \int \frac{dx}{f(x)} = C + \frac{x-x_0}{f(x_0)} - \frac{f'(x_0)}{2 f(x_0)^2} (x - x_0)^2 + \frac{2 f'(x_0)^2 - f''(x_0) f(x_0)}{6 f(x_0)^3} (x - x_0)^3 \cr + \frac{6 f''(x_0) f'(x_0) f(x_0) - f'''(x_0) f(x_0)^2 - 6 f'(x_0)^3}{24 f(x_0)^4} (x - x_0)^4 + \ldots\cr}$$

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    $\begingroup$ In fact, any $g(x)$ can be (formally) expanded as a series in terms of $f(x)$; this is the (Lagrange-)Bürmann series. $\endgroup$ Sep 6, 2011 at 23:42
  • $\begingroup$ @Robert Curioser and curioser. $\endgroup$
    – Pedro
    Mar 18, 2012 at 0:12

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