Answer
$$\int \frac{1}{1+e^{x}}dx=x-ln(e^{x}+1)+C$$
Work Step by Step
$$\int \frac{1}{1+e^{x}}dx=\int \frac{e^{-x}}{e^{-x}+1}dx$$
$$=-\int \frac{1}{e^{-x}+1}d(e^{-x}+1)=-ln(e^{-x}+1)+C$$
$$=-ln(\frac{e^{x}+1}{e^{x}})+C=x-ln(e^{x}+1)+C$$
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