For $$x>0,$$ let $$f\left( x \right) = \int\limits_e^x {{{\ln t} \over {1 + t}}dt.} $$ Find the function
$$f\left( x \right) + f\left( {{1 \over x}} \right)$$ and show that $$f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$$
Here, $$\ln t = {\log _e}t$$.