$\int \frac{e^{- x}}{1 + e^{- x}} d x$ $=$ $- p log (1 + e^{- x}) + C$ then p =

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Solution

$L e t 1 + e^{- x} = t ∴ - e^{- x} d x = d t ∴ \int (\frac{1}{t}) d t = - l o g t + C = - l o g (1 + e^{- x}) + C$

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