## University Calculus: Early Transcendentals (3rd Edition)

$-\ln (e^{-x}+1)+c$ $=x-\ln(1+e^x)+c$
Solve $\int \dfrac{dx}{1+e^x}$ This can be written as:$\int \dfrac{e^{-x}dx}{e^{-x}+1}$ Let us $p=e^{-x}+1$ and $dp=-e^{-p}dx$ This implies $\int \dfrac{e^{-x}dx}{e^{-x}+1} =-\int\dfrac{1}{p}dp$ $=-\ln |p|+c$ $=-\ln (e^{-x}+1)+c$ $=\ln (\dfrac{e^x}{1+e^x})+c$ Need to use logarithmic property, such as: $\ln m-\ln n=\ln (\dfrac{m}{n})$ Thus, $=\ln (e^x)-\ln(1+e^x)+c$ Hence, $=x-\ln(1+e^x)+c$