Question

Qual a integral de:

$\int\limits^2_0 \frac{e^x}{1+e^x} dx$

Answer 1

$\mathrm{Calculando\ \int\dfrac{e^x}{1+e^x}\ dx:}\\\\ \mathrm{u=1+e^x\ \ \| \ \ \mathrm{\dfrac{du}{dx}}=e^x\ \to\ du=e^x.dx}\\\\ \mathrm{\int\dfrac{1}{1+e^x}\ .\ e^x\ dx=\int\dfrac{1}{u}\ du=\ln{u}+C}\\\\ \mathbf{=\ln{(1+e^x)}+C}\\\\ \mathrm{Calculando\ \int\limits_0^2 \dfrac{e^x}{1+e^x}\ dx:}\\\\ \mathrm{\int\limits_0^2 \dfrac{e^x}{1+e^x}\ dx=\ln{(1+e^x)}\mid_0^2\ =}\\\\ \mathrm{=\ln{(1+e^2)-\ln{(1+e^0)}}=\ln{(1+e^2})-\ln{2}=}\\\\ \mathbf{=\ln{\bigg(\dfrac{1+e^2}{2}\bigg)}}$