Let `I_(1)=int_(0)^(alpha)(1+2cosx)/(1+e^(x))dx` and `I_(2)=int_(0)^(alpha)(1+e^(x))/(1+2cosx)dx`, where `alpha` is the root of the equation `2 cos x - e^(x)=0`. and `alpha` is positive Then,

Options

A. `I_(1)=I_(2)`
B. `I_(1) gt I_(2)`
C. `I_(1)+I_(2)=0`
D. `I_(1) ltI_(2)`

Answer

To solve the problem, we need to analyze the integrals \( I_1 \) and \( I_2 \) and compare their values.

### Step-by-Step Solution:

1. **Define the Integrals**:
\[ I_1 = \int_0^{\alpha} \frac{1 + 2 \cos x}{1 + e^x} \, dx \]
\[ I_2 = \int_0^{\alpha} \frac{1 + e^x}{1 + 2 \cos x} \, dx \]

2. **Identify the Value of \( \alpha \)**:
The value of \( \alpha \) is defined as the positive root of the equation:
\[ 2 \cos x - e^x = 0 \]
This implies:
\[ 2 \cos x = e^x \]

3. **Graphical Interpretation**:
To understand the behavior of the functions, we can graph \( y = 2 \cos x \) and \( y = e^x \). The intersection point where \( 2 \cos x = e^x \) gives us the value of \( \alpha \).

4. **Behavior of Functions**:
- For \( x \in [0, \alpha] \), since \( \alpha \) is the point where \( 2 \cos x \) meets \( e^x \), we can observe that \( 2 \cos x \) is greater than \( e^x \) for \( x \) in this interval.
- This implies:
\[ 1 + 2 \cos x > 1 + e^x \]

5. **Comparison of Integrands**:
Since \( 1 + 2 \cos x > 1 + e^x \), we can conclude:
\[ \frac{1 + 2 \cos x}{1 + e^x} > 1 \]
Therefore, \( I_1 \) is expected to be greater than the integral of 1 over the same interval.

6. **Evaluate \( I_2 \)**:
Conversely, since \( 1 + e^x > 1 + 2 \cos x \) in the interval, we have:
\[ \frac{1 + e^x}{1 + 2 \cos x} < 1 \]
Thus, \( I_2 \) is expected to be less than the integral of 1 over the same interval.

7. **Conclusion**:
From the analysis, we can conclude:
\[ I_1 > 1 \quad \text{and} \quad I_2 < 1 \]
Therefore, we can state:
\[ I_1 > I_2 \]

### Final Answer:
Thus, the answer is \( I_1 > I_2 \).