\displaystyle[br]\begin{equation*}\int \frac{\,\mathrm{d}p}{p(100-p)} = kt + c \iff \frac{1}{100} \int \frac{1}{p} - \frac{1}{p-100} \, \mathrm{d}p = kt + c\end{equation*}
\displaystyle[br]\begin{equation*}\int \frac{\,\mathrm{d}p}{p(100-p)} = kt + c \iff \frac{1}{100} \int \frac{1}{p} - \frac{1}{p-100} \, \mathrm{d}p = kt + c\end{equation*}
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