MathematicsEasy210×since 2002Q3456The 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 toA1112+loge4{{11} \over {12}} + {\log _e}41211+loge4B116+loge4{{11} \over 6} + {\log _e}4611+loge4C1112−loge4{{11} \over {12}} - {\log _e}41211−loge4D116−loge4{{11} \over 6} - {\log _e}4611−loge4Check answerSkip