Réponse : Bonjour,
[tex]I=\int_{0}^{1} \frac{e^{2x}}{1+e^{2x}} dx=[\frac{1}{2}\ln(1+e^{2x})]_{0}^{1}=\frac{1}{2}\ln(1+e^{2})-\frac{1}{2}\ln(1+e^{0})=\frac{1}{2}(\ln(1+e^{2})-\ln(2))=\frac{1}{2}\ln(\frac{1+e^{2}}{2}) \\I+J=\int_{0}^{1} \frac{e^{2x}+1}{1+e^{2x}} dx=\int_{0}^{1} 1 dx=[x]_{0}^{1}=1-0=1\\J=1-I=1-\frac{1}{2}\ln(\frac{1+e^{2}}{2})[/tex].
