For x > 0, if $f(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt}$, then $f(e) + f\left( {{1 \over e}} \right)$ is equal to :

$f(x) = \int_1^x {{{\ln t} \over {1 + t}}dt}$<br><br>then $f\left( {{1 \over x}} \right) = \int_1^{1/x} {{{\ln t} \over {1 + t}}dt}$<br><br>Let $t = {1 \over u} \Rightarrow dt = - {1 \over {{u^2}}}du$<br><br>$$ \Rightarrow f\left( {{1 \over x}} \right) = \int_1^x {{{\ln {1 \over u}} \over {1 + {1 \over u}}}\left( { - {1 \over {{u^2}}}} \right)dx} $$<br><br>$$f\left( {{1 \over x}} \right) = \int_1^x {{{\ln u} \over {u(1 + u)}}} du = \int_1^x {{{\ln t} \over {t(1 + t)}}dt} $$<br><br>$\therefore$ $$f(x) + f\left( {{1 \over x}} \right) = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over {t(1 + t)}}} \right)} dt$$<br><br>$$ = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over t} - {1 \over {t + 1}}} \right)} dt$$<br><br>$= \int_1^x {{{\ln t} \over t}dt = {1 \over 2}{{(\ln x)}^2}}$<br><br>$\therefore$ $f(e) + f\left( {{1 \over e}} \right) = {1 \over 2}{(\ln e)^2} = {1 \over 2}$