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If $f(x)$ is integrable and for all $x \in \mathbb{R}$, $f(x) \ne 0$, then is $1/f(x)$ integrable ?

I tried to prove that by Riemann or by Darboux's theorem but I couldn’t prove that.

LL 3.14
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Amin bsoul
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    I hate to nitpick: the adjective is integrable (as in integrable function) and it's Riemann, not "reman." – Sean Roberson May 18 '23 at 15:58
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    Under these conditions -- no. Take $f(x) = x$ if $x \neq 0$ and $f(0) = 1$. Then $1/f$ is unbounded on $[0,1]$ and not Riemann integrable. If you specify that $0 < m \leqslant f(x) \leqslant M$ everywhere then this has been answered already here. – RRL May 18 '23 at 16:28
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    ... and here – RRL May 18 '23 at 16:31

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