How do you find int(e^-x)/(1+e^-x)dx?

Jun 28, 2018

$\int {e}^{- x} / \left({e}^{- x} + 1\right) \mathrm{dx} = x - \ln \left({e}^{x} + 1\right) + c$

Explanation:

$\int {e}^{- x} / \left({e}^{- x} + 1\right) \mathrm{dx} = \int \frac{- \left({e}^{- x} + 1\right) '}{{e}^{- x} + 1} \mathrm{dx} = - \int \frac{\left({e}^{- x} + 1\right) '}{{e}^{- x} + 1} \mathrm{dx} = - \ln \left({e}^{- x} + 1\right) + c$

• $\ln \left({e}^{- x} + 1\right) = \ln \left(\frac{1}{e} ^ x + 1\right) = \ln \left(\frac{{e}^{x} + 1}{e} ^ x\right) = \ln \left({e}^{x} + 1\right) - \ln {e}^{x} = \ln \left({e}^{x} + 1\right) - x$

Therefore, $\int {e}^{- x} / \left({e}^{- x} + 1\right) \mathrm{dx} = - \left(\ln \left({e}^{x} + 1\right) - x\right) + c = x - \ln \left({e}^{x} + 1\right) + c$

, $c$$\in$$\mathbb{R}$

Jun 28, 2018

The answer is $= - \ln \left(1 + {e}^{-} x\right) + C$

Explanation:

Perform this integral by substitution

Let $u = 1 + {e}^{-} x$

$\implies$, $\mathrm{du} = - {e}^{-} x \mathrm{dx}$

Therefore, the integral is

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

$= - \int \frac{\mathrm{du}}{u}$

$= - \ln \left(u\right)$

$= - \ln \left(1 + {e}^{-} x\right) + C$

Jun 28, 2018

$- \ln \left(1 + {e}^{- x}\right) + C$

Explanation:

Let $u = {e}^{- x}$. This yields $\mathrm{du} = - {e}^{- x} \mathrm{dx}$

Rewrite

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

Substitute $- {e}^{- x} \mathrm{dx} = \mathrm{du}$ and ${e}^{- x} = u$:

$- \setminus \int \setminus \frac{1}{1 + u} \mathrm{du}$

This integral is $- \ln \left(1 + u\right) + C$.

Substitute back $u = {e}^{- x}$ to get

$\setminus \int \setminus \frac{{e}^{- x}}{1 + {e}^{- x}} \mathrm{dx} = - \ln \left(1 + {e}^{- x}\right) + C$

Jun 28, 2018

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