Question

Int(1 \(1 + {e}^{x} )​

Answer 1

Answer:

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

Step-by-step explanation:

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

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

In other words,

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

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

Let $u = 1 + e^{x}$. $u > 1$.

$du = e^{x}\cdot dx$.

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

Therefore

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