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Excuse me if this is a silly question, but this has never been proven to me and I cannot at the moment see a way to prove this to myself.

So the question is, Are integrals unique, beyond an integration constant? That is, is it true that if

$\int f(x)dx = F(x) + c$ and $\int f(x)dx = G(x) +c $

Then we have $F(x) = G(x)$, for all $x$?

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    $\begingroup$ Subtract the two equations and see what you get? $\endgroup$ – Chee Han Oct 12 '17 at 4:18
  • $\begingroup$ Actually yeah,thats very simple,I should just have sat down and had a look, thanks $\endgroup$ – Jandré Snyman Oct 12 '17 at 4:32
  • $\begingroup$ What do you mean? $\endgroup$ – Jandré Snyman Oct 12 '17 at 5:09
  • $\begingroup$ In this case I Am referring to indefinite integrals $\endgroup$ – Jandré Snyman Oct 12 '17 at 5:10
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Yes, this is a restatement of the fact $$f^{\prime}=0\implies f=\text{const.}$$ Which is a straightforward consequence of the mean value theorem.

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This is actually Quite Simple, Thanks to the user Chee Han for the hint at a solution tactic.

If $\int f(x)dx = F(x)+c$ and $\int f(x)dx = G(x)+c$ then we simply subtract the two equations to find

$\int f(x)dx -\int f(x)dx =F(x)+c-G(x)-c\\ \implies 0 =F(x)+c-G(x)-c\\ \implies F(X) = G(x)$

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