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December 3, 2025

log|x| + C revisited

Posted by Mike Shulman

A while ago on this blog, Tom posted a question about teaching calculus: what do you tell students the value of ∫1xdx\displaystyle\int \frac{1}{x}\,dx is? The standard answer is ln|x|+C\ln{|x|}+C, with CC an “arbitrary constant”. But that’s wrong if ∫\displaystyle\int means (as we also usually tell students it does) the “most general antiderivative”, since

F(x)={ln|x|+C − ifx<0 ln|x|+C + ifx>0 F(x) = \begin{cases} \ln{|x|} + C^- &\text{if}\;x\lt 0\\ \ln{|x|} + C^+ &\text{if}\;x\gt 0 \end{cases}

is a more general antiderivative, for two arbitrary constants C −C^- and C +C^+. (I’m writing ln\ln for the natural logarithm function that Tom wrote as log\log, for reasons that will become clear later.)

In the ensuing discussion it was mentioned that other standard indefinite integrals like ∫1x 2dx=−1x+C\displaystyle\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C are just as wrong. This happens whenever the domain of the integrand is disconnected: the “arbitrary constant” CC is really only locally constant. Moreover, Mark Meckes pointed out that believing in such formulas can lead to mistaken calculations such as

∫ −1 11x 2dx=−1x] −1 1=−2 \int_{-1}^1 \frac{1}{x^2}\,dx = \left.-\frac{1}{x}\right]_{-1}^1 = -2

which is “clearly nonsense” since the integrand is everywhere positive.

In this post I want to argue that there’s actually a very natural perspective from which ∫1x 2dx=−1x+C\displaystyle\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C is correct, while ∫1xdx=ln|x|+C\displaystyle\int \frac{1}{x}\,dx = \ln{|x|}+C is wrong for a different reason.

Continue reading “log|x| + C revisited” …
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