The value of $$\int\limits_0^1 {\frac{{dx}}{{{e^x} + e}}} $$   is equal to :

Solution :
$$\eqalign{ & {\text{Let }}I = \int\limits_0^1 {\frac{{dx}}{{{e^x} + e}}} = \int\limits_0^1 {\frac{{{e^x}dx}}{{{e^x}\left( {{e^x} + e} \right)}}} \cr & {\text{Put }}{e^x} = t \Rightarrow {e^x}dx = dt \cr & I = \int\limits_1^e {\frac{{dt}}{{t\left( {t + e} \right)}}} \cr & = \frac{1}{e}\int\limits_1^e {\left( {\frac{1}{t} - \frac{1}{{t + e}}} \right)} \cr & = \frac{1}{e}\int\limits_1^e {\frac{1}{t}dt} - \frac{1}{e}\int\limits_1^e {\frac{1}{{t + e}}dt} \cr & = \frac{1}{e}\left[ {\log \,t} \right]_1^e - \frac{1}{e}\left[ {\log \left( {t + e} \right)} \right]_1^e \cr & = \frac{1}{e}\left[ {\log \,t - \log \left( {t + e} \right)} \right]_1^e \cr & = \frac{1}{e}\left[ {\log \left( {\frac{t}{{t + e}}} \right)} \right]_1^e \cr & = \frac{1}{e}\left[ {\log \left( {\frac{e}{{2e}}} \right) - \log \left( {\frac{1}{{1 + e}}} \right)} \right] \cr & = \frac{1}{e}\log \left[ {\frac{{\frac{1}{2}}}{{\frac{1}{{\left( {1 + e} \right)}}}}} \right] \cr & = \frac{1}{e}\log \left( {\frac{{1 + e}}{2}} \right) \cr} $$