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

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

Vedclass · Accepted Answer

$x - \log(e^x + 1) + c$,where $c$ is the constant of integration.

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(B) To evaluate the integral $I = \int \frac{1}{e^x + 1} \, dx$,we can multiply the numerator and the denominator by $e^{-x}$:$I = \int \frac{e^{-x}}{e^{-x}(e^x + 1)} \, dx = \int \frac{e^{-x}}{1 + e^{-x}} \, dx$Let $u = 1 + e^{-x}$. Then $du = -e^{-x} \, dx$,which implies $e^{-x} \, dx = -du$.Substituting these into the integral:$I = \int \frac{-du}{u} = -\log|u| + c$$I = -\log(1 + e^{-x}) + c$Since $1 + e^{-x} = 1 + \frac{1}{e^x} = \frac{e^x + 1}{e^x}$,we have:$I = -\log\left(\frac{e^x + 1}{e^x}\right) + c = -[\log(e^x + 1) - \log(e^x)] + c$$I = -\log(e^x + 1) + x + c$Thus,$I = x - \log(e^x + 1) + c$.

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

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