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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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up vote 10 down vote accepted

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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In fact, any $g(x)$ can be (formally) expanded as a series in terms of $f(x)$; this is the (Lagrange-)Bürmann series. – J. M. Sep 6 '11 at 23:42
@Robert Curioser and curioser. – Pedro Tamaroff Mar 18 '12 at 0:12

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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For a somewhat simpler example along the same lines, take $f = \log$. – lhf Sep 6 '11 at 16:16
@lhf: Simpler in form only, since a change variables gives $$ \int\frac{\mathrm{d}x}{\log(x)}=\int\frac{\exp(u)\;\mathrm{d}u}{u} $$ – robjohn Sep 6 '11 at 16:30
@robjohn: hence "along the same lines" methinks. ;) – J. M. Sep 6 '11 at 16:32
@J. M.: fair enough. – robjohn Sep 6 '11 at 16:35

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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