\(\begin{align*} \int \sqrt{1+e^x}dx \end{align*}\) |
Connaissances:
- Changement de variable
- Astuce du -1/+1
- Décomposition en éléments simples
\(\begin{align*} I & = \int \sqrt{1+e^x}dx \end{align*}\)
Posons le changement de variable:
\(\begin{align*} & \begin{cases} u = \sqrt{1+e^x} \Rightarrow e^x = u²-1 \\
e^x dx =2u.du \Rightarrow dx = \frac{2u.du}{e^x} = \frac{2u}{u²-1}du \end{cases} \\
I & = \int u \times \frac{2u}{u²-1}du\\
& = 2 \int \frac{u²}{u²-1}du =2 \int \frac{u²-1+1}{u²-1}du\\
& = 2 \int du + 2\int\frac{1}{u²-1}du = \\
& = 2 \int du + 2\int\frac{1}{(u-1)(u+1)}du \\ \end{align*}\)
Décomposition en éléments simples:
\(\begin{align*} & \frac{1}{(u-1)(u+1)} = \frac{a}{u-1} + \frac{b}{u+1} \\ \\
& \begin{cases} \times (u-1) \text{ et }u=1 & \Rightarrow & a=1/2 \\
\times (u+1) \text{ et }u=-1 & \Rightarrow & b=-1/2
\end{cases} \\ \\
I & = 2 \int du + 2\int\frac{1}{(u-1)(u+1)}du \\
& = 2 \int du + \cancel{2}\int\frac{1/\cancel{2}}{(u-1)}du -\cancel{2} \int \frac{1/\cancel{2}}{u+1}du \\
& = 2u+ln(u-1) - ln(u+1) +C \\
& = 2\sqrt{1+e^x}+ln(\sqrt{1+e^x}-1) - ln(\sqrt{1+e^x}+1) +C \\
& = 2\sqrt{1+e^x}+ln \bigg(\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\bigg)+C \end{align*}\)
\[\boxed {\begin{align*} I & = 2\sqrt{1+e^x}+ln \bigg(\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\bigg)+C(\in \mathbb R) \end{align*}}\]