JEE Main 202325 Jan, Evening shiftMathematicsEasyIntegral Calculus369× since 2002The integral 16∫12dxx3(x2+2)216\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}}161∫2x3(x2+2)2dx is equal toAOption A. 1112+loge4{{11} \over {12}} + {\log _e}41211+loge4BOption B. 116+loge4{{11} \over 6} + {\log _e}4611+loge4COption C. 1112−loge4{{11} \over {12}} - {\log _e}41211−loge4DOption D. 116−loge4{{11} \over 6} - {\log _e}4611−loge4CheckSkip · show solution
