[tex]\ln(1+e^{nx}) = \ln\Big(e^{nx}(e^{-nx}+1)\Big) = \ln e^{nx}+\ln(1+e^{-nx})\\ \\ \displaystyle \lim\limits_{n\to \infty}\dfrac{1}{n}\int_{0}^1\ln(1+e^{nx})\, dx =\\ \\ = \lim\limits_{n\to \infty}\dfrac{1}{n}\int_{0}^1\ln e^{nx}\, dx +\lim\limits_{n\to \infty}\dfrac{1}{n}\int_{0}^1\ln(1+e^{-nx})\, dx =[/tex]
[tex]\displaystyle= \mathcal{L}_1+\mathcal{L}_2 \\ \\ \\\mathcal{L}_1 =\lim\limits_{n\to\infty}\dfrac{1}{n}\int_{0}^1 nx\ln e\, dx= \int_{0}^1 x\, dx = \dfrac{1}{2} \\ \\\\\mathcal{L}_2 = \lim\limits_{n\to \infty}\int_{0}^1\dfrac{\ln(1+e^{-nx})}{n}\, dx \\\\ \\0 <\dfrac{\ln(1+e^{-nx})}{n}\leq \dfrac{\ln 2}{n}\,\,\text{ (deoarece pentru }x=0\text{ are valoare maxima.)} \\ \\\\\Rightarrow0< \mathcal{L}_2 \leq 0 \Rightarrow \mathcal{L}_2 = 0\\ \\ \\ \Rightarrow \lim\limits_{n\to \infty}\dfrac{1}{n}\int_{0}^1\ln(1+e^{nx})\, dx = \dfrac{1}{2}+0 = \boxed{\dfrac{1}{2}}[/tex]
