If `I=int_-1^1 ([x^2]+log((2+x)/(2-x)))dx` where denotes the greatest integer `leq x`, then `I` equals

Options

A. `-2`
B. `-1`
C. 0
D. 1

Answer

To solve the integral \( I = \int_{-1}^{1} \left( [x^2] + \log\left(\frac{2+x}{2-x}\right) \right) dx \), where \([x^2]\) denotes the greatest integer function, we will break it down into two parts: \( I_1 \) and \( I_2 \).

### Step 1: Evaluate \( I_1 = \int_{-1}^{1} [x^2] \, dx \)

1. **Understanding the greatest integer function**:
- For \( x \in [-1, 1] \), \( x^2 \) varies from \( 0 \) to \( 1 \).
- Therefore, \( [x^2] = 0 \) for \( x \in [-1, 1) \) and \( [x^2] = 0 \) for \( x = 1 \) as well.

2. **Calculating \( I_1 \)**:
\[ I_1 = \int_{-1}^{1} [x^2] \, dx = \int_{-1}^{1} 0 \, dx = 0 \]

### Step 2: Evaluate \( I_2 = \int_{-1}^{1} \log\left(\frac{2+x}{2-x}\right) \, dx \)

1. **Using the property of definite integrals**:
- We will check the symmetry of the function \( f(x) = \log\left(\frac{2+x}{2-x}\right) \).
- Calculate \( f(-x) \):
\[ f(-x) = \log\left(\frac{2-x}{2+x}\right) = -\log\left(\frac{2+x}{2-x}\right) = -f(x) \]
- Since \( f(-x) = -f(x) \), the function is odd.

2. **Calculating \( I_2 \)**:
\[ I_2 = \int_{-1}^{1} f(x) \, dx = 0 \quad \text{(since the integral of an odd function over a symmetric interval is zero)} \]

### Step 3: Combine results to find \( I \)

1. **Combine \( I_1 \) and \( I_2 \)**:
\[ I = I_1 + I_2 = 0 + 0 = 0 \]

### Final Answer:
\[ I = 0 \]

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