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Just for amusement, not a question because the answer is not too difficult.

My old math teacher showed me more than 50 years ago how to integrate $\frac{1}{x}$ by parts. And this returns an unexpected result. $$\int\frac{1}{x}dx=\frac{1}{x}\cdot x-\int x d \frac{1}{x}=1+\int x\frac{1}{x^2}dx=1+\int\frac{1}{x}dx$$

Since $\int\frac{1}{x}dx=1+\int\frac{1}{x}dx$, 0 must equal 1

I reward my students for providing a good explanation. It explains a lot about integration by parts and finite/infinite integrals.

Paul vdVeen
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    Antiderivatives are not functions but equivalence classes of functions up to constant on connected parts. And $0$ is indeed equal to $1$ up to constant. – Carla only proves trivial prop Sep 19 '22 at 21:18
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    So, what did you want to do with this? – Mark Viola Sep 19 '22 at 21:18
  • That is the $f=x$ special case of this, which I'm sure is also discussed elsewhere. Since the aim here is - I think - to get students thinking rather than ask what's going on, I won't argue this is a duplicate. But if students struggling want an explanation, they should follow that link (plus any other someone finds for $f=x$) rather than post a new question. – J.G. Sep 19 '22 at 21:20
  • https://math.stackexchange.com/a/348395/471884 – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 19 '22 at 21:22
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    $\int 0\ dx$ is not just $0$, but $C$. – peterwhy Sep 19 '22 at 21:24
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    Implicitly, it is $1\bigg|_a^b=0$, which is not unexpected. – Quanto Sep 19 '22 at 21:36
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    This is an interesting way to get students think deeply about the intricacies of the indefinite integral, but these sorts of posts are usualy reserved for questions rather than exposition. You could consider editing this to ask about other similar tricks to get students seriously thinking about certain topics. That's a topic where there would certainly be interesting discussion. – Christian E. Ramirez Sep 19 '22 at 23:02
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    @C-RAM. "This is an interesting way to get students think deeply about the intricacies of the indefinite integral". I fully agree. Which was the only reason I posted it. This simple incorrect reasoning explains a lot about definite/indefinite integrals. Of course all comments makes sense. Maybe I should have posted it into another topic, more concerned about teaching mathematics. – Paul vdVeen Sep 20 '22 at 22:37

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