If $I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}} $, then :

Question

${1 \over 16} < {I^2} < {1 \over 9}$

${1 \over 8} < {I^2} < {1 \over 4}$

${1 \over 9} < {I^2} < {1 \over 8}$

${1 \over 6} < {I^2} < {1 \over 2}$

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