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 can use a substitution method and properties of logarithms. Here’s a step-by-step solution:

### Step 1: Substitute the variable
Let \( t = 1 + e^x \). Then, we differentiate to find \( dx \):
\[ \frac{dt}{dx} = e^x \implies dx = \frac{dt}{e^x} \]
From the substitution, we also have:
\[ e^x = t - 1 \]
Thus, we can rewrite \( dx \) as:
\[ dx = \frac{dt}{t - 1} \]

### Step 2: Change the limits of integration
When \( x = -a \):
\[ t = 1 + e^{-a} \]
When \( x = a \):
\[ t = 1 + e^{a} \]
So the integral becomes:
\[ I = \int_{1 + e^{-a}}^{1 + e^{a}} \frac{t - 1}{t} \cdot \frac{dt}{t - 1} \]
This simplifies to:
\[ I = \int_{1 + e^{-a}}^{1 + e^{a}} \frac{1}{t} \, dt - \int_{1 + e^{-a}}^{1 + e^{a}} \frac{1}{t} \, dt \]

### Step 3: Evaluate the integral
The integral can be split into two parts:
\[ I = \int_{1 + e^{-a}}^{1 + e^{a}} \frac{1}{t} \, dt - \int_{1 + e^{-a}}^{1 + e^{a}} \frac{1}{t} \, dt \]
Using the property of logarithms, we have:
\[ I = \left[ \ln t \right]_{1 + e^{-a}}^{1 + e^{a}} = \ln(1 + e^{a}) - \ln(1 + e^{-a}) \]

### Step 4: Simplify using logarithmic properties
Using the property of logarithms:
\[ I = \ln \left( \frac{1 + e^{a}}{1 + e^{-a}} \right) \]
We can rewrite \( 1 + e^{-a} \) as \( \frac{e^{a} + 1}{e^{a}} \):
\[ I = \ln \left( \frac{(1 + e^{a}) e^{a}}{1 + e^{a}} \right) = \ln(e^{a}) = a \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{a} \]