If $$I = \int_1^2 \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}}$$, then

We have to estimate the value of the definite integral

$$I \;=\;\int_{1}^{2}\dfrac{dx}{\sqrt{2x^{3}-9x^{2}+12x+4}}.$$

Put

$$g(x)=2x^{3}-9x^{2}+12x+4.$$

Then the integrand is $$f(x)=\dfrac{1}{\sqrt{g(x)}}.$$

To find upper and lower bounds for $$f(x)$$ on the interval $$[1,2]$$, we first study the behaviour of $$g(x)$$ itself.

Differentiate:

$$g'(x)=\dfrac{d}{dx}\bigl(2x^{3}-9x^{2}+12x+4\bigr)=6x^{2}-18x+12.$$

Factorising the quadratic part,

$$6x^{2}-18x+12 \;=\;6\bigl(x^{2}-3x+2\bigr) \;=\;6\bigl(x-1\bigr)\bigl(x-2\bigr).$$

This sign-analysis immediately follows:

• For $$1<x<2$$, the product $$\bigl(x-1\bigr)\bigl(x-2\bigr)<0$$, hence $$g'(x)<0.$$
• Exactly at the end-points, $$x=1$$ and $$x=2$$, we have $$g'(1)=0$$ and $$g'(2)=0.$$

So $$g(x)$$ decreases strictly on the open interval $$(1,2)$$ and attains its maximum at $$x=1$$ and minimum at $$x=2.$$ Now evaluate $$g(x)$$ at these two points:

At $$x=1:$$

$$g(1)=2(1)^{3}-9(1)^{2}+12(1)+4 =2-9+12+4 =9.$$

At $$x=2:$$

$$g(2)=2(2)^{3}-9(2)^{2}+12(2)+4 =16-36+24+4 =8.$$

Because $$g(x)$$ is decreasing, for every $$x\in[1,2]$$ we must have

$$8 \;\le\; g(x) \;\le\; 9.$$

Taking positive square roots preserves the inequality (all numbers involved are positive):

$$\sqrt{8} \;\le\; \sqrt{g(x)} \;\le\; \sqrt{9}.$$

Now reciprocate each side. Since each term is positive, the inequality sign reverses:

$$\dfrac{1}{\sqrt{9}} \;\le\; \dfrac{1}{\sqrt{g(x)}} \;\le\; \dfrac{1}{\sqrt{8}}.$$

That is, for all $$x \in [1,2]$$ we have the bound

$$\dfrac{1}{3} \;\le\; f(x) \;\le\; \dfrac{1}{2\sqrt{2}}.$$

Because the length of the integration interval is $$2-1=1,$$ we integrate these constant bounds term-by-term:

$$\int_{1}^{2}\dfrac{1}{3}\,dx \;\le\; I \;\le\; \int_{1}^{2}\dfrac{1}{2\sqrt{2}}\,dx.$$

Evaluating the two elementary integrals gives

$$\left.\dfrac{x}{3}\right|_{1}^{2} \;\le\; I \;\le\; \left.\dfrac{x}{2\sqrt{2}}\right|_{1}^{2},$$

so

$$\dfrac{2-1}{3} \;\le\; I \;\le\; \dfrac{2-1}{2\sqrt{2}}.$$

Simplify each side:

$$\dfrac{1}{3} \;\le\; I \;\le\; \dfrac{1}{2\sqrt{2}}.$$

We now square every term (all are positive, so inequality directions are preserved):

$$\left(\dfrac{1}{3}\right)^{2} \;<\; I^{2} \;<\; \left(\dfrac{1}{2\sqrt{2}}\right)^{2}.$$

Compute the squares:

$$\dfrac{1}{9} \;<\; I^{2} \;<\; \dfrac{1}{8}.$$

Thus $$I^{2}$$ lies strictly between $$\dfrac{1}{9}$$ and $$\dfrac{1}{8},$$ which coincides exactly with Option B.

Hence, the correct answer is Option 2.