`"I f"I_1=int_(-100)^(101)(dx)/((5+2x-2x^2)(1+e^(2-4x)))` `"and"I_2=int_(-100)^(101)(dx)/(5+2x-2x^2),t h e n(I_1)/(I_2)"i s"` 2 (b) `1/2` (c) 1 (d) `-1/2`

Options

A. `2`
B. `1/2`
C. `1`
D. `-1/2`

Explanation

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and find the ratio \( \frac{I_1}{I_2} \).

### Step 1: Define the integrals
We have:
\[ I_1 = \int_{-100}^{101} \frac{dx}{(5 + 2x - 2x^2)(1 + e^{2 - 4x})} \]
\[ I_2 = \int_{-100}^{101} \frac{dx}{5 + 2x - 2x^2} \]

### Step 2: Simplifying \( I_1 \)
To simplify \( I_1 \), we can use a substitution. Let's consider the substitution \( x = 1 - t \). Then \( dx = -dt \), and the limits change as follows:
- When \( x = -100 \), \( t = 101 \)
- When \( x = 101 \), \( t = -100 \)

Thus, we can rewrite \( I_1 \):
\[ I_1 = \int_{101}^{-100} \frac{-dt}{(5 + 2(1 - t) - 2(1 - t)^2)(1 + e^{2 - 4(1 - t)})} \]

### Step 3: Evaluate the denominator of \( I_1 \)
Now we simplify the denominator:
\[ 5 + 2(1 - t) - 2(1 - t)^2 = 5 + 2 - 2t - 2(1 - 2t + t^2) = 5 + 2 - 2t - 2 + 4t - 2t^2 = 3 + 2t - 2t^2 \]
And for the exponential term:
\[ 1 + e^{2 - 4(1 - t)} = 1 + e^{2 - 4 + 4t} = 1 + e^{-2 + 4t} \]

### Step 4: Rewrite \( I_1 \)
Now we can rewrite \( I_1 \):
\[ I_1 = \int_{-100}^{101} \frac{dt}{(3 + 2t - 2t^2)(1 + e^{-2 + 4t})} \]

### Step 5: Combine \( I_1 \) and \( I_2 \)
Notice that:
\[ I_1 + I_1 = \int_{-100}^{101} \left( \frac{1}{(5 + 2x - 2x^2)(1 + e^{2 - 4x})} + \frac{1}{(3 + 2t - 2t^2)(1 + e^{-2 + 4t})} \right) dx \]
This means:
\[ 2I_1 = I_2 \]

### Step 6: Find the ratio \( \frac{I_1}{I_2} \)
From the equation \( 2I_1 = I_2 \), we can express \( I_1 \) in terms of \( I_2 \):
\[ I_1 = \frac{1}{2} I_2 \]
Thus, the ratio is:
\[ \frac{I_1}{I_2} = \frac{1}{2} \]

### Final Answer
The answer is:
\[ \frac{I_1}{I_2} = \frac{1}{2} \]