$\int_{0}^{1}\frac{e^{x}}{e^{x}+e^{-x}}dx$
Ta có $I=\int_{0}^{1}\frac{e^x}{e^x+e^{-x}}dx=\int_{0}^{1}\frac{e^{2x}}{e^{2x}+1}dx=\frac{1}{2}\int_{0}^{1}\frac{d(e^{2x}+1)}{e^{2x}+1}=\frac{1}{2}\ln |e^{2x}+1||_{0}^{1}$
$=\frac{1}{2}\ln \frac{e^2+1}{2}$
- ducnahasd yêu thích