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Suppose we have a piecewise-defined function $f$ such that $$x<0 \implies f(x) = \ln -x + c_1 \implies f'(x) = -\frac1{-x} = \frac1x$$ $$x>0 \implies f(x) = \ln x + c_2 \implies f'(x) = \frac1x$$

Then by the definition of antidifferentiation, $f(x)$ is a solution of $\int\frac1xdx$. However, in math class, what I have been told was that $\int \frac1x dx = \ln |x| +c$, with only one constant term. Is there in fact a constraint on the two constant terms such that they have to be equal, to result in the simplification to a single constant term, or is the one-constant solution just not general?

busukxuan
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  • In elementary calculus, one is taught that $f'(x)=0$ implies that $f$ is constant. Experienced mathematicians realise that $f$ is only locally constant; this distinction is considered too subtle for beginners. – Angina Seng Jul 10 '17 at 03:45
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    @LordSharktheUnknown Do you mean for some values of $x$ or do you mean for all $x$? Also how does that relate to this question? – busukxuan Jul 10 '17 at 03:47
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    https://golem.ph.utexas.edu/category/2012/03/reader_survey_logx_c.html – Hans Lundmark Jul 10 '17 at 06:33

1 Answers1

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The constant terms on the two sides of a discontinuity can be different. This is clear by looking at the graph: you can shift up and down the graphs of the two sides independently without affecting the derivative.

Ted
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