Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 25

$\ln (1+e^r)+c$

Solve $\int \dfrac{e^r}{1+e^r} dr $ Let us $p=1+e^r$ and $dp=e^{r}dr$ This implies $\int \dfrac{e^r}{1+e^r} dr=\int\dfrac{1}{p}dp$ $=\ln |p|+c$ $=\ln (1+e^r)+c$ Hence, $\int \dfrac{e^r}{1+e^r} dr=\ln (1+e^r)+c$