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

Mar 10, 2018

$\ln | \frac{1}{1 + {e}^{-} x} | + C .$

#### Explanation:

Subst. $1 + {e}^{-} x = t , \text{ so that, } - {e}^{-} x \mathrm{dx} = \mathrm{dt}$.

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

$= \int - \frac{1}{t} \mathrm{dt} = - \int \frac{1}{t} \mathrm{dt}$,

$= - \ln | t |$,

$= \ln | t {|}^{-} 1$,

$\Rightarrow I = \ln | \frac{1}{1 + {e}^{-} x} | + C .$

Mar 10, 2018

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

#### Explanation:

Perform the substitution

$u = 1 + {e}^{-} x$, $\implies$, $\mathrm{du} = - {e}^{-} x \mathrm{dx}$

Therefore,

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

$= - \ln u$

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

Mar 10, 2018

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

#### Explanation:

$I = \int \frac{{e}^{-} x}{1 + {e}^{-} x} \mathrm{dx}$,
take, ${e}^{-} x = t \implies {e}^{-} x \left(- 1\right) \mathrm{dx} = \mathrm{dt} \implies {e}^{-} x \mathrm{dx} = - \mathrm{dt}$
$I = - \int \frac{\mathrm{dt}}{1 + t} = - \ln | 1 + t | + c = - \ln | 1 + {e}^{-} x | + c$
$I = \ln | \frac{1}{1 + {e}^{-} x} | + c$