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How to find all functions $g: \mathbb{R} \to \mathbb{R}$ such that for all $x \neq 0$,

$$g'(x) = \frac{1}{x^2}$$

I know that the antiderivative is $-1/x$, but how to find all $g$?

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  • $\begingroup$ Do you have a good first guess at a solution? (Do you know any functions with derivative $1/x^2$?) $\endgroup$ Nov 30 '15 at 16:59
  • $\begingroup$ @MarioCarneiro f(x) = - 1/x $\endgroup$
    – Kittu
    Nov 30 '15 at 17:07
  • $\begingroup$ If $g(x)$ is a solution, then $(g-f)'(x)=0$ everywhere except $0$, so $g-f$ is locally a constant on $\Bbb R^{>0}$, $\Bbb R^{<0}$, and $\{0\}$. Thus the solutions enumerated by copper.hat are the only ones. $\endgroup$ Nov 30 '15 at 17:10
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There are three sets to consider, $x <0$, $\{0\}$ and $x>0$. The first and last are connected and $g'$ is continuous there, hence we can write $g(x) = g(1)+ \int_1^x g'(t)dt$ for $x >0$ and $g(x) = g(-1)+ \int_{-1}^x g'(t)dt$. We can choose $g(0)$ arbitrarily.

Hence we need to choose constants $g(-1), g(0)$ and $g(1)$, then we have $g(x) = \begin{cases} g(-1)- ({1 \over x} +1), & x <0 \\ g(0), & x = 0 \\ g(1)-({1 \over x}-1), & x > 0 \end{cases}$

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