\(\begin{align*} \int \frac{1}{1+e^x}dx \end{align*}\) |
Connaissances:
- Méthode 1:
- astuce de calcul (+1/-1)
- Reconnaissance de la forme \(u'/u\)
- Méthode 2:
- Changement de variable
- Décomposition en éléments simples
Methode 1:
\(\begin{align*} I & = \int \frac{1}{1+e^x}dx \\
& = \int \frac{1+e^x-e^x}{1+e^x}dx \\
& = \int \frac{1+e^x}{1+e^x}dx- \int \frac{e^x}{1+e^x}dx \\
& = \int dx- \int \frac{e^x}{1+e^x}dx \\
& = \boxed{x-ln(1+e^x) + C} \end{align*}\)
Méthode 2:
Posons: \(u=1+e^x\)
\( \Rightarrow \begin{cases} e^x = u-1 \\ e^xdx =du \Rightarrow dx = \frac{du}{e^x}= \frac{du}{u-1}\end{cases} \)
\(\begin{align*} I & = \int \frac{1}{1+e^x}dx \\
& = \int \frac{1}{u(u-1)}du \end{align*}\)
Décomposition en éléments simples:
\(\begin{align*} & \frac{1}{u(u-1)} = \frac{a}{u} + \frac{b}{u-1} \\
& \begin{cases} \times (u-1) \text{ et }u=1 & \Rightarrow & b=1 \\
\text {si } u =-1 \Rightarrow 1/2 = -a-1/2 & \Rightarrow & a=-1
\end{cases} \\ \\
I & = \int -\frac{1}{u}du + \int \frac{1}{u-1}du \\
& = -lnu + ln(u-1) +C \\
& = -ln(1+e^x) + ln(\cancel{1}+e^x-\cancel{1}) +C \\
& = \boxed{x-ln(1+e^x) + C} \end{align*}\)
\[\boxed {\begin{align*} I & = x-ln(1+e^x) + C(\in \mathbb R) \end{align*}}\]