MathematicsMedium210×since 2002Q3399For x > 0, if f(x)=∫1xloget(1+t)dtf(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt}f(x)=1∫x(1+t)logetdt, then f(e)+f(1e)f(e) + f\left( {{1 \over e}} \right)f(e)+f(e1) is equal to :A12{1 \over 2}21B−-−1C0D1Check answerSkip