= Solution
A suitable real <antiderivative> on the disconnected domain is
$$
\boxed{G(x)=\ln|x|,\qquad x\ne0.}
$$
For $x>0$, $G'=1/x$ directly; for $x<0$, $G=\ln(-x)$ and the <chain rule> again gives $G'=1/x$. The graph is even, crosses zero at $x=\pm1$, tends to $-\infty$ on either side of zero, and tends to $+\infty$ as $|x|\to\infty$. Independent additive constants may be chosen on the two disconnected half-lines; both are set to zero here.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1-q17-antiderivative-branches.png]
{title=Real <antiderivative> branches for $1/x$ and $1/\sqrt{x^2-1}$, with their domains and limiting behavior}
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