1. We need to evaluate $I=\int_1^2 \frac{dx}{\sqrt{2x^3-9x^2+12x+4}}.$

2. First simplify the cubic inside the square root.

Let $f(x)=2x^3-9x^2+12x+4.$ Check whether it factors: $f(2)=16-36+24+4=8,$ $f(1)=2-9+12+4=9.$ Try factoring using $(x-2)$: $2x^3-9x^2+12x+4=(x-2)(2x^2-5x-2).$ Now factor the quadratic: $2x^2-5x-2=(x-\tfrac{1}{2})(2x+4)?$ That is not convenient. Instead, observe directly: $2x^3-9x^2+12x+4=(x-2)^2(2x+1).$ Indeed, $(x-2)^2(2x+1)=(x^2-4x+4)(2x+1)=2x^3-8x^2+8x+x^2-4x+4=2x^3-7x^2+4x+4,$ which is not correct, so re-check.

Let us factor carefully: Since $f(2)=0$ is false, $(x-2)$ is not a factor. So instead use a substitution approach.

3. Notice that $f'(x)=6x^2-18x+12=6(x-1)(x-2).$ This suggests shifting around $x=\tfrac32$. Let $x=t+\frac32.$ Then $f(x)=2\left(t+\frac32\right)^3-9\left(t+\frac32\right)^2+12\left(t+\frac32\right)+4.$ Expanding, $f(x)=2t^3-\frac32 t+\frac{17}{4}.$ This is not especially helpful.

4. A better idea is to inspect whether the integrand can be estimated on $[1,2]$. Since $f(1)=9,\qquad f(2)=8,$ and $f'(x)=6(x-1)(x-2),$ for $1\le x\le 2$, we have $(x-1)(x-2)\le 0 \implies f'(x)\le 0.$ So $f(x)$ is decreasing on $[1,2]$. Hence $8\le f(x)\le 9.$ Therefore, $\frac1{\sqrt9}\le \frac1{\sqrt{f(x)}}\le \frac1{\sqrt8},$ that is, $\frac13\le \frac1{\sqrt{2x^3-9x^2+12x+4}}\le \frac1{2\sqrt2}.$ Integrating from $1$ to $2$ (interval length $=1$), $\frac13<I<\frac1{2\sqrt2}.$

5. Now square the inequality. Since all terms are positive, $\frac19<I^2<\frac18.$

6. Compare with the options:

• A: $\frac1{16}<I^2<\frac19$ — false
• B: $\frac18<I^2<\frac14$ — false
• C: $\frac19<I^2<\frac18$ — true
• D: $\frac16<I^2<\frac12$ — false

Therefore, the correct option is $\boxed{\text{C}}.$