
\begin{enumerate}
\item On peut écrire $f(x)$ sous la forme:
\[
f(x) = \frac{1}{1+\e^{-x}} = \frac{1+\e^{-x}-\e^{-x}}{1+\e^{-x}} = 1-\frac{\e^{-x}}{1+\e^{-x}} = 1+ \frac{u'(x)}{u(x)}, \text{ avec }u(x)=1+\e^{-x}.
\]
Par conséquent,
\begin{align*}
\mathcal{A} & =\int_{-5}^2 \left(1 + \frac{-\e^{-x}}{1+\e^{-x}} \right)\dx\\
            & = \left[ x + \ln\left(1+\e^{-x} \right) \right]_{-5}^2\\
            & = 2+\ln\big(1+\e^{-2}\big) - (-5) - \ln\big(1+\e^5 \big)\\
          & = 7 + \ln\left( \frac{1+\e^{-2}}{1+\e^5}\right) \\
          & = 7 + \ln\left( \frac{1+\tfrac{1}{\e^2}}{1+\e^5} \right)\\
          & = 7 + \ln\left( \frac{\tfrac{\e^2+1}{\e^2}}{1+\e^5} \right)\\
          & = 7 + \ln\left( \dfrac{1}{\e^2}\times \dfrac{\e^2+1}{1+\e^5} \right)\\
          & = 7 + \underbrace{\ln\left(\frac{1}{\e^2}\right)}_{=-2}+ \ln\left( \frac{\e^2+1}{1+\e^5} \right)\\
\Aboxed{\mathcal{A} & = 5 + \ln\left( \frac{1+\e^2}{1+\e^5} \right)}
\end{align*}

\item On a:
\begin{align*}
\mathcal{A}(t) & = \int_t^2 f(x) \dx\\
               & = \left[ x + \ln\big(1+\e^{-x}\big) \right]_t^2\\
               & = 2+\ln\big(1+\e^{-2}\big) - t - \ln\big(1+\e^{-t}\big)\\
            & = (2-t) + \ln\left( \frac{1+\e^{-2}}{1+\e^{-t}} \right) \\
        & = (2-t) + \ln\left( \frac{\tfrac{\e^2+1}{\e^2}}{\tfrac{\e^t+1}{\e^t}} \right)   \\
         & = (2-t) + \ln\left( \frac{\e^t}{\e^2} \times \frac{1+\e^2}{1+\e^t} \right)\\
         & = (2-t) + \ln\frac{\e^t}{\e^2} + \ln\left( \frac{1+\e^2}{1+\e^t} \right)\\
         & = (2-t) + t-2 + \ln\left( \frac{1+\e^2}{1+\e^t} \right)\\
\Aboxed{\mathcal{A}(t) & = \ln\left( \frac{1+\e^2}{1+\e^t} \right)}
\end{align*}

$\limite{t}{-\infty}{\big( \e^t \big) }=0$ donc:
\[
\boxed{ \limite{t}{-\infty}{\mathcal{A}(t)} = \ln\big(1+\e^2\big) }
\]
\end{enumerate}
\vspace*{2cm}
