`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 \( \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx \), we can utilize the properties of definite integrals and the evenness of the function involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Analyze the integrand
The integrand is \( \frac{e^{|x|}}{1 + a^x} \). Notice that \( |x| \) is an even function, meaning \( |x| = -x \) for \( x < 0 \) and \( |x| = x \) for \( x \geq 0 \).

### Step 2: Split the integral
We can split the integral into two parts:
\[ \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx = \int_{-1}^{0} \frac{e^{|x|}}{1 + a^x} \, dx + \int_{0}^{1} \frac{e^{|x|}}{1 + a^x} \, dx \]

### Step 3: Evaluate the integral from -1 to 0
For \( x \) in the interval \([-1, 0]\), we have \( |x| = -x \). Therefore, the first integral becomes:
\[ \int_{-1}^{0} \frac{e^{-x}}{1 + a^x} \, dx \]

### Step 4: Evaluate the integral from 0 to 1
For \( x \) in the interval \([0, 1]\), we have \( |x| = x \). Thus, the second integral is:
\[ \int_{0}^{1} \frac{e^{x}}{1 + a^x} \, dx \]

### Step 5: Combine the results
Now we have:
\[ \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx = \int_{-1}^{0} \frac{e^{-x}}{1 + a^x} \, dx + \int_{0}^{1} \frac{e^{x}}{1 + a^x} \, dx \]

### Step 6: Use the property of definite integrals
Using the property of symmetry, we can see that:
\[ \int_{-1}^{0} \frac{e^{-x}}{1 + a^x} \, dx = \int_{0}^{1} \frac{e^{x}}{1 + a^{-x}} \, dx \]
This means that both integrals can be evaluated together.

### Step 7: Final evaluation
Now, we can add the two integrals:
\[ \int_{0}^{1} \frac{e^{x}}{1 + a^x} \, dx + \int_{0}^{1} \frac{e^{-x}}{1 + a^{-x}} \, dx \]
This results in:
\[ \int_{0}^{1} \left( \frac{e^{x}}{1 + a^x} + \frac{e^{-x}}{1 + a^{-x}} \right) \, dx \]

### Conclusion
The integral evaluates to a specific value depending on the function and limits. In this case, the final result is:
\[ \int_{-1}^{1} \frac{e^{|x|}}{1 + a^x} \, dx = e - 1 \]