Is it simple?

$since$

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

$therefore,$

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

$now\ we\ will\ make\ a\ substitution\ putting\ t\ =\ \frac { 1 }{ z }$

$hence\ dt\ =-\frac { 1 }{ { z }^{ 2 } } dz$

$and$ $f(\frac { 1 }{ x } )$ $transforms$ $to$ $\int _{ 1 }^{ x }{ \frac { log(\frac { 1 }{ z } ) }{ \frac { 1 }{ z } +1 } } (-\frac { 1 }{ { z }^{ 2 } } dz)$ $=\ \int _{ 1 }^{ x }{ \frac { log(z) }{ (z+1)z } } dz$

$and\ then\ replace\ z\ by\ t,$

$hence\ F(x)\ =\ f(x)+f(\frac { 1 }{ x } )$ = $\\ \int _{ 1 }^{ x }{ [\frac { log(t) }{ t+1 } dt } +\frac { log(t) }{ (t+1)t } ]dt\ =\ \int _{ 1 }^{ x }{ \frac { log(t) }{ t } } dt$ $\\hence\ F(e)\ =\ \int _{ 1 }^{ e }{ \frac { logt }{ t } } dt.$

$take\ log(t)$ = $'u'\ and\ solve\ the\ integral\ easily\ to\ obtain\ F(e)\ =\ 0.5$

$\begin{aligned} S & = f(e) + f(e^{-1}) \\ & = \int_1^e \frac {\log t}{1+t} dt + \color{#3D99F6} \int_1^\frac 1e \frac {\log t}{1+t}dt & \small \color{#3D99F6} \text{Let }u = \frac 1t \implies du = - \frac {dt}{t^2} \\ & = \int_1^e \frac {\log t}{1+t} dt + \color{#3D99F6} \int_1^e \frac {-\log \frac 1u}{u^2\left(1+\frac 1u\right)}du \\ & = \int_1^e \frac {\log t}{1+t} dt + \color{#3D99F6} \int_1^e \frac {\log u}{u(u+1)}du & \small \color{#3D99F6} \text{Replace } u \text{ with }t. \\ & = \int_1^e \left(\frac {\log t}{1+t} + \frac {\log t}{t(1+t)} \right) dt \\ & = \int_1^e \frac {t\log t + \log t}{t(1+t)} dt \\ & = \int_1^e \frac {\log t}t dt & \small \color{#3D99F6} \text{By integration by parts} \\ & = \log^2 t \bigg|_1^e - \color{#3D99F6} \int_1^e \frac {\log t}t dt & \small \color{#3D99F6} \text{Note that } \int_0^e \frac {\log t}t dt = S \\ & = \frac {\log^2 e - \log^2 1}2 = \frac 12 = \boxed {0.5} \end{aligned}$