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

$F\left(e\right)=f\left(e\right)+f\left(\frac{1}{e}\right)$

$={\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln t}{1+t}dt}+{\displaystyle \underset{1}{\overset{1/e}{\int }}\frac{\ln t}{1+t}dt}$

In second integral substitute $t=\frac{1}{u}$

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

${\displaystyle \underset{1}{\overset{e}{\int }}\frac{\ln t}{t}}dt$

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