int_0^1 frac{e^{-x}}{1 + e^{-x}} ,dx = | vedclass

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(D) Let $I = \int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx$. Substitute $t = 1 + e^{-x}$,then $dt = -e^{-x} \,dx$,which implies $e^{-x} \,dx = -dt$. When $x

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(D) Let $I = \int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx$. Substitute $t = 1 + e^{-x}$,then $dt = -e^{-x} \,dx$,which implies $e^{-x} \,dx = -dt$. When $x = 0$,$t = 1 + e^0 = 2$. When $x = 1$,$t = 1 + e^{-1} = 1 + \frac{1}{e} = \frac{e+1}{e}$. Substituting these into the integral: $I = \int_2^{\frac{e+1}{e}} \frac{-dt}{t} = -[\log |t|]_2^{\frac{e+1}{e}}$. $I = -\left( \log \left( \frac{e+1}{e} \right) - \log 2 \right) = \log 2 - \log \left( \frac{e+1}{e} \right) = \log \left( \frac{2e}{e+1} \right)$. Wait,let us re-evaluate the original integral: $I = \int_0^1 \frac{1}{e^x + 1} \,dx = \int_0^1 \frac{e^x}{e^x(e^x+1)} \,dx$. Let $u = e^x$,$du = e^x \,dx$. $I = \int_1^e \frac{du}{u(u+1)} = \int_1^e \left( \frac{1}{u} - \frac{1}{u+1} \right) \,du = [\log u - \log(u+1)]_1^e = [\log \frac{u}{u+1}]_1^e = \log \frac{e}{e+1} - \log \frac{1}{2} = \log \frac{2e}{e+1}$. Since $\log \frac{2e}{e+1} = \log 2 + \log e - \log(e+1) = \log 2 + 1 - \log(e+1)$. Comparing with options,none match. Thus,the correct answer is $(D)$.

$\int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx = $

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