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If $$\int_{}^{} f(x)\,dx$$ is known, is there a way to directly find $$\int_{}^{} \frac{1}{f(x)}\,dx$$

Rory Daulton
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AHumbleCoder
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    Not really, no. Sorry. Take $f(x)=x$, then you get $x^2/2$ and $\ln x$, respectively. – vadim123 Mar 20 '15 at 15:38
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    For example, the integral of $\frac{x}{e^x}$, that is, of $xe^{-x}$, is nice and simple, but the integral of $\frac{e^x}{x}$ is not an elementary function. – André Nicolas Mar 20 '15 at 15:39
  • The motivation was very natural which puzzled me for long time to disprove. – MathArt Feb 28 '24 at 13:13

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Consider the function $f(x)$ in a power series. Let \begin{align} f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n} \end{align} then it can be seen that \begin{align} \frac{1}{f(x)} = \frac{1}{a_{0}} - \frac{a_{1}}{a_{0}^{2}} \, x + \frac{a_{1}^{2} - a_{2}}{a_{0}^{3}} \, x^{2} + \cdots \end{align} Integration of each shows \begin{align} \int f(x) \, dx &= a_{0} x + \frac{a_{1}}{2} \, x^{2} + \frac{a_{2}}{3} \, x^{3} + \cdots \\ \int \frac{dx}{f(x)} &= \frac{x}{a_{0}} - \frac{a_{1}}{2 \, a_{0}^{2}} \, x^{2} + \frac{a_{1}^{2} - a_{2}}{3 \, a_{0}^{3}} \, x^{3} + \cdots \end{align}

Leucippus
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    +1, nice answer, but one could argue if this is a way to "directly find" the integral. – Rory Daulton Mar 20 '15 at 17:10
  • @RoryDaulton Without knowing what $f(x)$ is it is hard to directly evaluate the integrals. The display given here is a way of how different the integral evaluations are. – Leucippus Mar 20 '15 at 17:15
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    You are right, of course, and it is good that you made the motivation explicit. As I said, nice answer! – Rory Daulton Mar 20 '15 at 17:18