Let $$F\left( x \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right),$$     where $$f\left( x \right) = \int\limits_l^x {\frac{{\log \,t}}{{1 + t}}dt.} $$    Then $$F\left( e \right)$$  equals :

Solution :
$$\eqalign{ & {\text{Given, }}F\left( x \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right), \cr & {\text{where }}f\left( x \right) = \int_1^x {\frac{{\log \,t}}{{1 + t}}dt} \cr & \therefore \,F\left( e \right) = f\left( e \right) + f\left( {\frac{1}{e}} \right) \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\log \,t}}{{1 + t}}dt} + \int_1^{\frac{1}{e}} {\frac{{\log \,t}}{{1 + t}}dt} ......\left( {\text{A}} \right) \cr & {\text{Now for solving, }}I = \int_1^{\frac{1}{e}} {\frac{{\log \,t}}{{1 + t}}dt} \cr & \therefore \,{\text{Put }}\frac{1}{t} = z \Rightarrow - \frac{1}{{{t^2}}}dt = dz \Rightarrow dt = - \frac{{dz}}{{{z^2}}} \cr & {\text{and limit for }}t = 1 \Rightarrow z = 1{\text{ and for }}t = \frac{1}{e} \Rightarrow z = e \cr & \therefore \,I = \int_1^e {\frac{{\log \left( {\frac{1}{z}} \right)}}{{1 + \frac{1}{z}}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & \Rightarrow I = \int_1^e {\frac{{\left( {\log \,1 - \log \,z} \right).z}}{{z + 1}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & \Rightarrow I = \int_1^e { - \frac{{\log \,z}}{{\left( {z + 1} \right)}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & \Rightarrow I = \int_1^e {\frac{{\log \,z}}{{z\left( {z + 1} \right)}}dz} \cr & \therefore \,I = \int_1^e {\frac{{\log \,t}}{{t\left( {t + 1} \right)}}dt} \cr & {\text{Equation}}\left( {\text{A}} \right){\text{ becomes :}} \cr & F\left( e \right) = \int_1^e {\frac{{\log \,t}}{{1 + t}}dt + \int_1^e {\frac{{\log \,t}}{{t\left( {1 + t} \right)}}dt} } \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{t.\log \,t + \log \,t}}{{t\left( {1 + t} \right)}}dt} \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\left( {\log \,t} \right)\left( {t + 1} \right)}}{{t\left( {1 + t} \right)}}dt} \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\log \,t}}{t}dt} \cr & {\text{Let }}\log \,t = x \cr & \therefore \,\frac{1}{t}dt = dx \cr & \left[ {{\text{for limit }}t = 1,{\text{ }}x = 0{\text{ and }}t = e,\,x = \log \,e = 1} \right] \cr & \therefore \,F\left( e \right) = \int_0^1 {x\,dx} \cr & \Rightarrow F\left( e \right) = \left[ {\frac{{{x^2}}}{2}} \right]_0^1 \cr & \Rightarrow F\left( e \right) = \frac{1}{2} \cr} $$