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

$F\left(e\right)=f\left(e\right)+f\left(\frac{1}{e}\right)={\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln t}{1+t}dt+{\displaystyle \underset{1}{\overset{\frac{1}{e}}{\int }}\frac{\ln t}{1+t}dt}}$

Setting $t=\frac{1}{u}$  in the second integral,

$F\left(e\right)={\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln t}{1+t}}dt+{\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln u}{u\left(1+u\right)}}du$

           $={\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln t}{1+t}}\left(1+\frac{1}{t}\right)dt={\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln t}{t}}dt$

           $={\frac{{\left(\ln t\right)}^2}{2}|}_1^e=\frac{1}{2}$