In finding this anti-derivative: $$\int\frac{1}{f(x)}\,dx,$$ with $f(x) \not \neq0$ not everywhere, i.e., there is $x_0$ such that $f(x_0)=0$;

would it be possible to prove that the anti-derivative must contain a logarithmic term, i.e., either: $$\ln(f(x)) \text{ or } \log(f(x)) \text{?}$$

I don't know if this proposition is even true, but the result of a problem that I am working on implies that this must be the case.

Thank you.

  • $\begingroup$ Do you mean f'/f instead of 1/f? $\endgroup$ – randomgirl Apr 10 '17 at 4:00
  • $\begingroup$ What if $f(x) = {1 \over x}$? $\endgroup$ – copper.hat Apr 10 '17 at 4:00
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    $\begingroup$ Let $f(x)= \sqrt x$ for $x\ge 0.$ Any antiderivative for $1/f$ on $(0,\infty)$ has the form $2\sqrt x + C.$ $\endgroup$ – zhw. Apr 10 '17 at 5:49

Suppose $f(x) = 1+x^2.$ Then $$ \int \frac 1 {f(x)} \, dx = \int \frac 1 {1+x^2} \, dx = \arctan x + C. $$ That is not $\log f(x).$

Note that in this context $\text{“}{\ln}\text{''}$ and $\text{“}{\log}\text{''}$ typically both mean the same thing, i.e. the logarithmic function whose base is $e\approx2.71828\ldots\,.$

  • $\begingroup$ Thank you a lot. Could you please look at my edited question in which I imposed a constraint on $f(x)$? $\endgroup$ – A Slow Learner Apr 10 '17 at 4:37
  • $\begingroup$ @Marmousi It is still false. Consider $f(x)=x^2$. $\endgroup$ – Olivier Apr 10 '17 at 21:34

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