`int_(-1)^(1) (e^(|x|))/(1+a^(x))dx`

Options

A. `e`
B. `2-e`
C. `e-3`
D. `e-1`

Explanation

To solve the integral \( I = \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx \), we will use the property of definite integrals and the symmetry of the function involved.

### Step-by-Step Solution:

1. **Recognize the Integral**:
We start with the integral:
\[ I = \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx \]

2. **Use the Property of Definite Integrals**:
We can use the property that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
Here, \( a = -1 \) and \( b = 1 \). Thus:
\[ I = \int_{-1}^{1} \frac{e^{|-x|}}{1 + a^{-x}} \, dx \]
Since \( |-x| = |x| \), we can rewrite this as:
\[ I = \int_{-1}^{1} \frac{e^{|x|}}{1 + \frac{1}{a^x}} \, dx \]

3. **Simplify the Integral**:
The term \( 1 + \frac{1}{a^x} \) can be rewritten as:
\[ 1 + \frac{1}{a^x} = \frac{a^x + 1}{a^x} \]
Therefore, we can express the integral as:
\[ I = \int_{-1}^{1} \frac{e^{|x|} a^x}{a^x + 1} \, dx \]

4. **Combine the Two Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx \quad \text{and} \quad I = \int_{-1}^{1} \frac{e^{|x|} a^x}{a^x + 1} \, dx \]
Adding these two equations gives:
\[ 2I = \int_{-1}^{1} e^{|x|} \, dx \]

5. **Evaluate the Integral**:
Now, we need to evaluate \( \int_{-1}^{1} e^{|x|} \, dx \). Since \( |x| \) is an even function, we can simplify this:
\[ \int_{-1}^{1} e^{|x|} \, dx = 2 \int_{0}^{1} e^{x} \, dx \]

6. **Calculate the Integral from 0 to 1**:
The integral \( \int e^{x} \, dx \) is \( e^{x} \). Therefore:
\[ \int_{0}^{1} e^{x} \, dx = e^{1} - e^{0} = e - 1 \]

7. **Final Calculation**:
Substituting back, we have:
\[ 2I = 2(e - 1) \implies I = e - 1 \]

### Final Answer:
\[ I = e - 1 \]