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I've read in a book that for a central force of the form $$ f(r)= \frac{{-ke^ {-r/a}}}{r^2} $$ the adequate potential is $$ V(r)= \frac{{-ke^ {-r/a}}}{r} $$

I'm trying to understand why $$ -\frac{\partial{V}}{\partial{r}}=f(r) $$ I've tried checking in wolfram and some other sites, but still can't figure it out. I thought perhaps there's some assumption being made that eliminates the second element in the derivative of V, but can't think of something. can someone please help me understand how to integrate this thing?

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    $\begingroup$ A strange thing here is that $f(r) \neq -V'(r)$. $\endgroup$ – Gec Jan 13 '20 at 17:18
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    $\begingroup$ may be they mean k'=k-1/a for the potential $\endgroup$ – Wolphram jonny Jan 13 '20 at 18:04
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    $\begingroup$ What book? Are you certain that you have included all of the contextual assumptions? $\endgroup$ – Bill N Jan 13 '20 at 18:05
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    $\begingroup$ The fact that they refer to the potential as being adequate suggests to me that in the context where you see this formula used, the “a” in the denominator of the second term of the derivative is very large compared with the “r”, so that the second term is negligible compared with the first. $\endgroup$ – R.W. Bird Jan 13 '20 at 18:21
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    $\begingroup$ Which book? Which page? $\endgroup$ – Qmechanic Jan 13 '20 at 19:30