JEE 2007

$\begin{aligned} f(x) & = \int_1^x \frac {\log t}{1+t} dt & \small \color{#3D99F6} \text{Let }t = e^u \implies dt = e^u du \\ & = \int_0^{\log x} \frac {ue^u}{1+e^u} du \\ \implies f\left(\frac 1x\right) & = \int_0^{\color{#3D99F6} \log \frac 1x} \frac {ue^u}{1+e^u} du \\ & = \int_0^{\color{#3D99F6}-\log x} \frac {ue^u}{1+e^u} du & \small \color{#3D99F6} \text{Let }u=-v \implies du = - dv \\ & = \int_0^{\log x} \frac {ve^{-v}}{1+e^{-v}} dv & \small \color{#3D99F6} \text{Multiply up and down by }e^v \\ & = \int_0^{\log x} \frac {v}{e^v + 1} dv \end{aligned}$

Therefore,

$\begin{aligned} F(x) & = f(x) + f\left(\frac 1x\right) \\ & = \int_0^{\log x} \frac {ue^u}{1+e^u} du + \int_0^{\log x} \frac u{1+e^u} du \\ & = \int_0^{\log x} u \ du = \frac {\log^2 x}2 \\ \implies F(e) & = \frac {\log^2 e}2 = \boxed{\dfrac 12} \end{aligned}$

Observe that the substitution $t = 1/u$ leads us to

$f \left( \frac{1}{x} \right) = \int_1^{\frac{1}{x}} \frac{\log t}{t+1} dt = \int_1^x \frac{\log \left( 1/u \right)}{\frac{1}{u} + 1} \left( \frac{-1}{u^2} \right ) du \overset{u \rightarrow t}{=} \int_1^{x} \frac{\log t}{t(t+1)} dt,$

and therefore

$F(x) = f(x) + f\left(\frac{1}{x}\right) = \int_1^{x} \left(\frac{\log t}{(t+1)} + \frac{\log t}{t(t+1)} \right )dt = \int_1^x \frac{\log t}{t} dt.$

Here integration by parts will show us that

$F(x) = \frac{\log^2 x}{2},$

and therefore $F(e) = 1/2.$