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Recently I found the following "proof" that 0=1 online, but I don't see where it is wrong. You start with the following integral:

$$\int \frac{1}{x} dx $$ You can direcly solve it to $ln(x)+k$, but let's say we do it by parts. $$ \int u·dv = u·v-\int v·du$$ Where $u=\frac{1}{x}$, $du=-\frac{1}{x^2}dx$, $dv=dx$ and $v=x$: $$\int \frac{1}{x} dx=\frac{1}{x}·x-\int x·\frac{-1}{x^2}dx$$ Simplifying we get: $$\int \frac{1}{x} dx = 1+\int \frac{1}{x} dx$$ $$\int \frac{1}{x} dx - \int \frac{1}{x} dx = 1$$ $$ 0=1 $$

Where did I mess up?

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    Indefinite integrals are only defined up to a constant. – Kavi Rama Murthy Feb 20 '22 at 09:18
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    Well, I firmly believe that looking at proofs where 0=1 is a waste of time, since we know that there is obviously something wrong with the reasoning. On the other hand, all the constants of the primitives I do not see. – A. P. Feb 20 '22 at 09:19
  • See also https://math.stackexchange.com/questions/3489112/c-in-integration-by-parts-allows-for-major-differences-in-answers and https://math.stackexchange.com/questions/2785130/are-the-following-equalities-true and https://math.stackexchange.com/questions/806254/using-integration-by-parts-results-in-0-1 – Gerry Myerson Feb 20 '22 at 09:25
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    It's worth nothing it works fine with definite integration. With $u=1/t,,v=t$ we have $\int_1^x\frac1tdt=\underbrace{[1]1^x}{0}+\int_1^x\frac1tdt$. – J.G. Feb 20 '22 at 09:25

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The fact that $$f'(x) = g'(x)$$ does not imply $$f(x) = g(x).$$ For instance, $$f(x) = x^2 + 1, \quad g(x) = x^2,$$ satisfies $f'(x) = g'(x) = 2x$. Why would you think that an antiderivative of a function is unique?

heropup
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