JEE Main 202126 Feb, Evening shiftMathematicsMediumIntegral Calculus369× since 2002For 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 :AOption A. 12{1 \over 2}21BOption B. −-−1COption C. 0DOption D. 1CheckSkip · show solution
