Let $F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right),$ where $f\left(x\right)=$$\underset{1}{\overset{x}{\int }}\frac{\ln t}{1+t}dt,$ then $F\left(e\right)=$

$\begin{array}{l}F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right)\\ f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\ln t}{1+t}dt\\ f\left(\frac{1}{x}\right)=\underset{1}{\overset{1/x}{\int }}\frac{\ln t}{1+t}dt\\ f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{\ln \left(\frac{1}{t}\right)}{1+\frac{1}{t}}d\left(\frac{1}{t}\right)\\ \left[\therefore \mathrm{Re}placingtby\frac{1}{t}\right]\end{array}$

$\begin{array}{l}=\underset{1}{\overset{x}{\int }}\frac{-\ln t}{\frac{t+1}{t}}\left(\frac{-1}{t^2}\right)dt\\ =\underset{1}{\overset{x}{\int }}\frac{\ln t}{t(t+1)}dt\end{array}$

$\begin{array}{l}Nowf\left(x\right)+f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}[\frac{\ln t}{1+t}+\frac{\ln t}{t(t+1)}]dt\\ =\underset{1}{\overset{x}{\int }}\frac{\ln t}{(1+t)}\left(1+\frac{1}{t}\right)dt\\ =\underset{1}{\overset{x}{\int }}\ln t\left(\frac{1}{t}\right)dt\\ Put\ln t=y\\ \frac{1}{t}dt=dy\end{array}$

$\begin{array}{l}=\underset{0}{\overset{\ln x}{\int }}y dy\\ ={{\left(\frac{y^2}{2}\right)}_0}^{\ln x}={\frac{\left(\ln x\right)}{2}}^2\\ F\left(e\right)=\frac{{\left(\ln e\right)}^2}{2}=\frac{1}{2}\end{array}$