int frac{dx}{1 + e^x} = | vedclass

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Question

(B) To evaluate the integral $I = \int \frac{dx}{1 + e^x}$,multiply the numerator and denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{e^{-x}(1 + e^x)}

Vedclass · Accepted Answer

$-\log(1 + e^{-x}) + C$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{dx}{1 + e^x}$,multiply the numerator and denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{e^{-x}(1 + e^x)} dx = \int \frac{e^{-x}}{e^{-x} + 1} dx$Let $t = 1 + e^{-x}$. Then,differentiating with respect to $x$,we get $dt = -e^{-x} dx$,which implies $e^{-x} dx = -dt$.Substituting these into the integral:$I = \int \frac{-dt}{t} = -\int \frac{1}{t} dt = -\log|t| + C$Substituting back $t = 1 + e^{-x}$:$I = -\log(1 + e^{-x}) + C$Thus,the correct option is $B$.

$\int \frac{dx}{1 + e^x} = $

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