The value of the integral `int_(-a)^(a)(e^(x))/(1+e^(x))dx` is

Options

A. `e^(a^(2))`
B. `a`
C. `e^(-a^(2))`
D. `(a)/(2)`

Answer

To solve the integral \( I = \int_{-a}^{a} \frac{e^x}{1 + e^x} \, dx \), we will use a substitution method and properties of logarithms.

### Step-by-Step Solution:

1. **Substitution**: Let \( t = 1 + e^x \). Then, we differentiate \( t \) with respect to \( x \):
\[ dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x} \]
From the substitution \( t = 1 + e^x \), we can express \( e^x \) as \( e^x = t - 1 \).

2. **Change of Limits**: When \( x = -a \):
\[ t = 1 + e^{-a} = 1 + \frac{1}{e^a} \]
When \( x = a \):
\[ t = 1 + e^{a} \]

3. **Transform the Integral**: The integral can be rewritten as:
\[ I = \int_{1 + \frac{1}{e^a}}^{1 + e^a} \frac{t - 1}{t} \cdot \frac{dt}{t - 1} \]
This simplifies to:
\[ I = \int_{1 + \frac{1}{e^a}}^{1 + e^a} \frac{dt}{t} = \int_{1 + \frac{1}{e^a}}^{1 + e^a} dt \]

4. **Evaluate the Integral**: The integral of \( \frac{1}{t} \) is \( \log(t) \). Thus, we have:
\[ I = \left[ \log(t) \right]_{1 + \frac{1}{e^a}}^{1 + e^a} \]
This gives:
\[ I = \log(1 + e^a) - \log\left(1 + \frac{1}{e^a}\right) \]

5. **Use Logarithmic Properties**: We can use the property of logarithms:
\[ \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \]
Therefore:
\[ I = \log\left(\frac{1 + e^a}{1 + \frac{1}{e^a}}\right) \]

6. **Simplify the Expression**: The term \( 1 + \frac{1}{e^a} \) can be rewritten as:
\[ 1 + \frac{1}{e^a} = \frac{e^a + 1}{e^a} \]
Thus:
\[ I = \log\left(\frac{(1 + e^a)e^a}{e^a + 1}\right) = \log(e^a) = a \]

### Final Result:
The value of the integral is:
\[ I = a \]