SigMathS
Réponse 42:
$$\begin{align*}
f(x)&=x+\int_{0}^{ln2}{f(y)e^x(e^y+e^{-y})dy}\\
&=x+e^x\int_{0}^{ln2}{f(y)(e^y+e^{-y})dy}\\
&=x+e^x(a+b) \\
\end{align*}$$
avec $a=\int_{0}^{ln2}{f(y)e^y\,dy}$ et $b=\int_{0}^{ln2}{f(y)e^{-y}\,dy}$
$$\begin{align*}
a&=\int_{0}^{ln2}{f(y)e^y dy}\\
&=\int_{0}^{ln2}{(y+e^y(a+b))e^ydy}\\
&=\int_{0}^{ln2}{ye^ydy}+(a+b)\int_{0}^{ln2}{e^{2y}dy}\\
&=\left[{ye^y}\right]_0^{ln2}-\int_{0}^{ln2}{e^ydy}+(a+b)\left[{\dfrac{1}{2}e^{2y}}\right]_0^{ln2}\\
&=2\ln 2-1+\dfrac{3}{2}(a+b) \qquad\qquad (i)\\
b&=\int_{0}^{ln2}{f(y)e^{-y} dy}\\
&=\int_{0}^{ln2}{(y+e^y(a+b))e^{-y}dy}\\
&=\int_{0}^{ln2}{ye^{-y}dy}+(a+b)\int_{0}^{ln2}{dy}\\
&=\left[{-ye^{-y}}\right]_{0}^{ln2}+\int_{0}^{ln2}{e^{-y}dy}+(a+b)ln2\\
&=(a+b)\ln 2+\dfrac{1-\ln 2}{2} \qquad\qquad (ii)
\end{align*}$$
$(i)$ et $(ii)\Longrightarrow$ $\left\{{\begin{aligned}&{a+3b=2-4\ln 2}\\&{2a\ln 2+2b(\ln 2-1=\ln 2-1)}\end{aligned}}\right.$
$\iff \left\{{\begin{aligned}&{(a+b)+2b=2-4\ln 2}\\&{2(a+b)\ln 2-2b=\ln 2-1}\end{aligned}}\right.$
Donc $a+b=\dfrac{1-3\ln 2}{1+2\ln 2}$
Alors $\boxed{f(x)=x+\dfrac{1-3\ln 2}{1+2\ln 2}e^x}$