For $x>0,$ if $f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{{\log }_{e}t}{(1+t)}dt,$ then $f\left(e\right)+f\left(\frac{1}{e}\right)$ is equal to 

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{x}\right)& =& {\int }_1^{\mathrm{x}} \frac{\log \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}\\ \mathrm{f}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\right)& =& {\int }_1^{1/\mathrm{x}} \frac{\log \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}\\ & & \mathrm{Replace} \mathrm{t} \mathrm{by} 1/\mathrm{t}\\ & =& {\int }_1^{\mathrm{x}} \frac{\log {\displaystyle \frac{1}{\mathrm{t}}}}{1+{\displaystyle \frac{1}{\mathrm{t}}}} \left(\frac{-1}{{\mathrm{t}}^2} \mathrm{dt}\right)\\ & =& {\int }_1^{\mathrm{x}} \frac{\log \mathrm{t}}{\mathrm{t} (\mathrm{t}+1)}\mathrm{dt}\end{array}$

Now

$$\begin{gathered} \begin{array}{rcl} & & f\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{x}\right) \\ ={\int }_1^{\mathrm{x}} \left[\frac{\log \mathrm{t}}{1+\mathrm{t}}+\frac{\log \mathrm{t}}{\mathrm{t}(\mathrm{t}+1)}\right] \mathrm{dt} \\ ={\int }_1^{\mathrm{x}} \frac{\log \mathrm{t}}{1+\mathrm{t}} \left(1+\frac{1}{\mathrm{t}}\right) \mathrm{dt} \\ ={\int }_1^{\mathrm{x}} \frac{\log \mathrm{t}}{\mathrm{t}} \mathrm{dt} \\ ={\left(\frac{{(\log \mathrm{t})}^2}{2}\right)}_1^x\\ & =& \frac{{(\log \mathrm{x})}^2}{2}\end{array} \end{gathered}$$

Now

$$\begin{gathered} \mathrm{f}\left(\mathrm{e}\right)+\mathrm{f}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{e}$}\right.\right) \\ =\frac{{\left(\log \mathrm{e}\right)}^2}{2} \\ =\frac{1}{2} \end{gathered}$$