Int(1 \(1 + {e}^{x} )

Question

asked Apr 20, 2020 184k views

1 Answer

Squaregoldfish · Answer 1 · 2020-04-27T05:31:11+0000

Answer:

$\begin{aligned}\int{(1)/(1 + e^(x))\cdot dx}= x - \ln(1 + e^(x)) + C\end{aligned}$ .

Explanation:

The first derivative of the denominator
1 + e^(x) is
e^(x) . Rewrite the fraction to obtain that expression on the numerator.

$\begin{aligned}(1)/(1 + e^(x)) &= (1 + e^(x))/(1 + e^(x)) - (e^(x))/(1+e^(x))\\&=1-(e^(x))/(1+e^(x))\end{aligned}$ .

In other words,

$\begin{aligned}\int{(1)/(1 + e^(x))\cdot dx} &= \int{dx} - \int{(e^(x))/(1+e^(x))\cdot dx}\end{aligned}$ .

Apply
-substitution on the integral
$\displaystyle \int{(e^(x))/(1+e^(x))\cdot dx}$ :

Let
u = 1 + e^(x) .
u > 1 .

$du = e^(x)\cdot dx$ .

$\displaystyle \int{(e^(x))/(1+e^(x))\cdot dx} = \int{(du)/(u)} = ln(|u|) = ln(u) +C = \ln{(1+e^(x))}+C$ .

Therefore

$\begin{aligned}\int{(1)/(1 + e^(x))\cdot dx} &= \int{dx} - \int{(e^(x))/(1+e^(x))\cdot dx}\\ & = x - \ln{(1 + e^(x))}+C\end{aligned}$ .