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),{\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} .....({\text{A}})\,} \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 & = \int_1^e {\frac{{\left( {\log \,1 - \log \,z} \right).z}}{{z + 1}}} \left( { - \frac{{dz}}{{{z^2}}}} \right) \cr & = \int_1^e { - \frac{{\log \,z}}{{\left( {z + 1} \right)}}\left( { - \frac{{dz}}{z}} \right)} \cr & = \int_1^e {\frac{{\log \,z}}{{z\left( {z + 1} \right)}}dz} \,\,\,\left[ {\therefore \log \,1 = 0} \right] \cr & \therefore I = \int_1^e {\frac{{\log \,t}}{{t\left( {t + 1} \right)}}dt} \cr & \left[ {{\text{By property }}\int_a^b {f\left( t \right)dt} = \int_a^b {f\left( x \right)dx} } \right] \cr & {\text{Equation (A) 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 & = \int_1^e {\frac{{t.\log \,t + \log \,t}}{{t\left( {1 + t} \right)}}dt} \cr & = \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 time }}t = 1,\,x = 0{\text{ and }}t = e,x = \log \,e = 1} \right] \cr & \therefore F\left( e \right) = \int_0^1 {x\,dx} \cr & F\left( e \right) = \left[ {\frac{{{x^2}}}{2}} \right]_0^1\, \Rightarrow F\left( e \right) = \frac{1}{2} \cr} $$