The value of `int_(-2)^(2)(3x^(2))/(1+e^(x))dx`, is

Options

A. 8
B. 2
C. 4
D. 0

Explanation

To solve the integral \( I = \int_{-2}^{2} \frac{3x^2}{1 + e^x} \, dx \), we will use the property of definite integrals.

### Step-by-Step Solution:

1. **Identify the Integral**:
\[ I = \int_{-2}^{2} \frac{3x^2}{1 + e^x} \, dx \]

2. **Use the Property of Definite Integrals**:
The property states that:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx \]
Here, we will find \( f(-x) \):
\[ f(-x) = \frac{3(-x)^2}{1 + e^{-x}} = \frac{3x^2}{1 + e^{-x}} \]

3. **Combine \( f(x) \) and \( f(-x) \)**:
Now we can write:
\[ I = \int_{0}^{2} \left( \frac{3x^2}{1 + e^x} + \frac{3x^2}{1 + e^{-x}} \right) \, dx \]

4. **Simplify the Expression**:
To combine the fractions:
\[ \frac{3x^2}{1 + e^x} + \frac{3x^2}{1 + e^{-x}} = 3x^2 \left( \frac{1 + e^{-x} + 1 + e^x}{(1 + e^x)(1 + e^{-x})} \right) \]
Simplifying the numerator:
\[ 1 + e^{-x} + 1 + e^x = 2 + e^x + e^{-x} = 2 + 2\cosh(x) \]
The denominator becomes:
\[ (1 + e^x)(1 + e^{-x}) = 1 + e^x + e^{-x} + 1 = 2 + 2\cosh(x) \]
Thus, we have:
\[ I = \int_{0}^{2} \frac{3x^2 (2 + 2\cosh(x))}{2 + 2\cosh(x)} \, dx = \int_{0}^{2} 3x^2 \, dx \]

5. **Evaluate the Integral**:
Now we can compute:
\[ I = 3 \int_{0}^{2} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from 0 to 2:
\[ \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]
Therefore:
\[ I = 3 \cdot \frac{8}{3} = 8 \]

### Final Answer:
The value of the integral is \( \boxed{8} \).