Exercice 35

\(\begin{align*} \int \frac{1}{e^x+e^{-x}}dx \end{align*}\)

Connaissances:

  • exponentielles
  • changement de variable
  • primitive de \(\frac{1}{x²+1}=tan^{-1}x\)

\(\begin{align*} I & = \int \frac{1}{e^x+e^{-x}}dx \\
& =\int \frac{e^x}{(e^x)²+1}dx \end{align*}\)

Posons le changement de variable \(u =e^x\) 

\(\begin{align*} & \begin{cases} u = e^x \\ du = e^x dx \end{cases} \\
I & =\int \frac{e^x}{(e^x)²+1}dx  \\
& =  \int \frac{du}{u²+1}dx \\
& = tan^{-1}u+C \\
& = tan^{-1}(e^x)+C
\end{align*}\)

\[ \boxed{\begin{align*} I = tan^{-1}(e^x)+C (\in \mathbb R)\end{align*}}\]