Let `I=int_1^2 (dx)/sqrt(2x^3-9x^2+12x+4)` then

Options

A. `1/9 lt I^2 lt 1/8`
B. `1/3 lt I^2 lt 1/2`
C. `1/9 lt I lt 1/8`
D. `1/3 lt I lt 1/2`

Explanation

To solve the integral \( I = \int_1^2 \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}} \), we will analyze the function inside the integral and estimate its value.

### Step 1: Define the function
Let \( f(x) = \frac{1}{\sqrt{2x^3 - 9x^2 + 12x + 4}} \).

### Step 2: Find the derivative
To analyze the behavior of \( f(x) \), we need to find its derivative \( f'(x) \):
\[ f'(x) = -\frac{1}{2} (2x^3 - 9x^2 + 12x + 4)^{-3/2} \cdot (6x^2 - 18x + 12) \]

### Step 3: Set the derivative to zero
Setting \( f'(x) = 0 \) gives us the critical points:
\[ 6x^2 - 18x + 12 = 0 \implies x^2 - 3x + 2 = 0 \]
Factoring, we find:
\[ (x - 1)(x - 2) = 0 \implies x = 1, 2 \]

### Step 4: Analyze the sign of the derivative
To determine the behavior of \( f(x) \) on the interval \( [1, 2] \):
- For \( x < 1 \), \( f'(x) > 0 \) (increasing)
- For \( x = 1 \), \( f'(x) = 0 \) (local minimum)
- For \( x = 2 \), \( f'(x) = 0 \) (local maximum)
- For \( x > 2 \), \( f'(x) < 0 \) (decreasing)

Thus, \( f(x) \) is decreasing on \( [1, 2] \).

### Step 5: Evaluate \( f(1) \) and \( f(2) \)
Calculating \( f(1) \):
\[ f(1) = \frac{1}{\sqrt{2(1)^3 - 9(1)^2 + 12(1) + 4}} = \frac{1}{\sqrt{2 - 9 + 12 + 4}} = \frac{1}{\sqrt{9}} = \frac{1}{3} \]

Calculating \( f(2) \):
\[ f(2) = \frac{1}{\sqrt{2(2)^3 - 9(2)^2 + 12(2) + 4}} = \frac{1}{\sqrt{16 - 36 + 24 + 4}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]

### Step 6: Estimate the integral
Since \( f(x) \) is decreasing, we can estimate the integral \( I \):
\[ I \text{ is bounded by } f(2) \cdot (2 - 1) \text{ and } f(1) \cdot (2 - 1) \]
Thus,
\[ \frac{1}{2\sqrt{2}} < I < \frac{1}{3} \]

### Step 7: Square the bounds
Squaring the bounds gives:
\[ \left(\frac{1}{2\sqrt{2}}\right)^2 < I^2 < \left(\frac{1}{3}\right)^2 \]
Calculating:
\[ \frac{1}{8} < I^2 < \frac{1}{9} \]

### Conclusion
The value of \( I^2 \) lies between \( \frac{1}{8} \) and \( \frac{1}{9} \).