For x > 0 , if f x = ∫ 1 x log e t 1 + t d t , then f e + f 1 e is equal to

$f\left(x\right)={\int }_1^x\frac{{\log }_{e}t}{\left(1+t\right)}dt$

$f\left(\frac{1}{x}\right)={\int }_1^{1/x}\frac{\ell nt}{1+t}dt$, let $t=\frac{1}{y}$

$={\int }_1^x\frac{\ell n\left(1/y\right)}{1+\left(1/y\right)}.\left(\frac{-1}{y^2}\right)dy$

$=+{\int }_1^x\frac{\ell ny}{1+y}.\frac{y}{y^2}dy$

$={\int }_1^x\frac{\ell ny}{y\left(1+y\right)}dy$

hence 

$f\left(x\right)+f\left(\frac{1}{x}\right)={\int }_1^x\frac{\left(1+t\right)\ell nt}{t\left(1+t\right)}dt={\int }_1^x\frac{\ell nt}{t}dt$

$=\frac{1}{2}{\ln }^2\left(x\right)$

so $f\left(e\right)+f\left(\frac{1}{e}\right)=\frac{1}{2} ...\left(3\right)$