I think this is a very silly question but I have some problems nonetheless.
If I know that $g'=-\frac{1}{f''}$, is then $$ g=(f')^{-1}? $$
2026-04-04 20:58:19.1775336299
Computing the integral of $-1/f''$
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There is no formula in general to get a closed form of $$ \int\frac1{f''} $$
We have $$ \left(\frac1f\right)'=-\frac{f'}{f^2} $$ and your expression for $g$ is not correct in general.