JEE Main 201912 Jan, Evening shiftMathematicsEasyIntegral Calculus369× since 2002The integral ∫1e{(xe)2x−(ex)x} \int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,1∫e{(ex)2x−(xe)x} log_e x dx is equal to :AOption A. −12+1e−12e2- {1 \over 2} + {1 \over e} - {1 \over {2{e^2}}}−21+e1−2e21BOption B. 32−e−12e2{3 \over 2} - e - {1 \over {2{e^2}}}23−e−2e21COption C. 12−e−1e2{1 \over 2} - e - {1 \over {{e^2}}}21−e−e21DOption D. 32−1e−12x2{3 \over 2} - {1 \over e} - {1 \over {2{x^2}}}23−e1−2x21CheckSkip · show solution
