MathematicsMedium210×since 2002Q3382If I=∫12dx2x3−9x2+12x+4I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}}I=1∫22x3−9x2+12x+4dx, then :A116<I2<19{1 \over 16} < {I^2} < {1 \over 9}161<I2<91B18<I2<14{1 \over 8} < {I^2} < {1 \over 4}81<I2<41C19<I2<18{1 \over 9} < {I^2} < {1 \over 8}91<I2<81D16<I2<12{1 \over 6} < {I^2} < {1 \over 2}61<I2<21Check answerSkip