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

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Question

(A) Let $I = \int \frac{e^{-x}}{1 + e^x} \, dx$. Multiply the numerator and denominator by $e^{-x}$: $I = \int \frac{e^{-x} \cdot e^{-x}}{e^{-x}(1 + e

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{e^{-x}}{1 + e^x} \, dx$. Multiply the numerator and denominator by $e^{-x}$: $I = \int \frac{e^{-x} \cdot e^{-x}}{e^{-x}(1 + e^x)} \, dx = \int \frac{e^{-2x}}{e^{-x} + 1} \, dx$. Let $t = e^{-x} + 1$. Then $dt = -e^{-x} \, dx$,which implies $e^{-x} \, dx = -dt$. Also,$e^{-x} = t - 1$. Substituting these into the integral: $I = \int \frac{t - 1}{t} (-dt) = -\int (1 - \frac{1}{t}) \, dt = \int (\frac{1}{t} - 1) \, dt$. Integrating,we get: $I = \log|t| - t + c$. Substituting $t = e^{-x} + 1$ back: $I = \log(e^{-x} + 1) - (e^{-x} + 1) + c$. Since $\log(e^{-x} + 1) = \log(\frac{1 + e^x}{e^x}) = \log(1 + e^x) - \log(e^x) = \log(1 + e^x) - x$,$I = \log(1 + e^x) - x - e^{-x} - 1 + c$. Since $-1 + c$ is just another constant $C$,the final answer is $\log(1 + e^x) - x - e^{-x} + C$.

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

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