Consider $$\int \frac{df(x)}{dx}dx$$

It's the same as $f(x)+c$ because we just differentiate and then integrate f(x).

But on the other hand, cancel out $dx$ gives $\int df(x)$

So $$f(x)+c=\int df(x)$$

Is this true? If so, what's $\int df(x)$?

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    $\begingroup$ It is an abuse of notation. I wouldn't read much into it to be honest. It's akin to cancelling differentials to motivate chain rule. It's a nice mnemonic device, but it is not great for actual understanding of the mathematics. $\endgroup$ – Cameron Williams Jun 13 '18 at 2:27
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    $\begingroup$ @CameronWilliams Riemann-Stieltjes Integral $\endgroup$ – Mark Viola Jun 13 '18 at 2:39
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    $\begingroup$ @MarkViola I don't entirely disagree, but like differential forms, I think it's a slight abuse of notation. It's an abuse that works, but is very prone to creating mistaken understandings. $\endgroup$ – Cameron Williams Jun 13 '18 at 2:50

set $y=f(x)$, we have $dy=f'(x)dx$, $$\int f'(x)dx=\int f'(x)\frac{dy}{f'(x)}=\int dy=y+c=f(x)+c$$ This is just the Variable transformation in integration.


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