`int_(-1)^(2)[([x])/(1+x^(2))]dx`, where [.] denotes the greatest integer function, is equal to

Options

A. `-2`
B. `-1`
C. zero
D. none of these

Explanation

To solve the definite integral \( I = \int_{-1}^{2} \frac{[x]}{1+x^2} \, dx \), where \([x]\) denotes the greatest integer function, we will break the integral into segments based on the behavior of the greatest integer function.

### Step 1: Identify the intervals for \([x]\)
The function \([x]\) (greatest integer function) takes constant integer values in the intervals:
- From \(-1\) to \(0\), \([x] = -1\)
- From \(0\) to \(1\), \([x] = 0\)
- From \(1\) to \(2\), \([x] = 1\)

### Step 2: Split the integral
We can split the integral \(I\) into three parts based on the intervals identified:
\[ I = \int_{-1}^{0} \frac{[x]}{1+x^2} \, dx + \int_{0}^{1} \frac{[x]}{1+x^2} \, dx + \int_{1}^{2} \frac{[x]}{1+x^2} \, dx \]

### Step 3: Evaluate each integral
1. **For the interval \([-1, 0]\)**:
\[ \int_{-1}^{0} \frac{[x]}{1+x^2} \, dx = \int_{-1}^{0} \frac{-1}{1+x^2} \, dx = -\int_{-1}^{0} \frac{1}{1+x^2} \, dx \]
The integral \(\int \frac{1}{1+x^2} \, dx\) gives \(\tan^{-1}(x)\):
\[ = -\left[ \tan^{-1}(x) \right]_{-1}^{0} = -\left( \tan^{-1}(0) - \tan^{-1}(-1) \right) = -\left( 0 - \left(-\frac{\pi}{4}\right) \right) = \frac{\pi}{4} \]

2. **For the interval \([0, 1]\)**:
\[ \int_{0}^{1} \frac{[x]}{1+x^2} \, dx = \int_{0}^{1} \frac{0}{1+x^2} \, dx = 0 \]

3. **For the interval \([1, 2]\)**:
\[ \int_{1}^{2} \frac{[x]}{1+x^2} \, dx = \int_{1}^{2} \frac{1}{1+x^2} \, dx \]
Again, using \(\int \frac{1}{1+x^2} \, dx = \tan^{-1}(x)\):
\[ = \left[ \tan^{-1}(x) \right]_{1}^{2} = \tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}(2) - \frac{\pi}{4} \]

### Step 4: Combine the results
Now, we can combine the results from the three intervals:
\[ I = \frac{\pi}{4} + 0 + \left( \tan^{-1}(2) - \frac{\pi}{4} \right) \]
This simplifies to:
\[ I = \tan^{-1}(2) \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-1}^{2} \frac{[x]}{1+x^2} \, dx = \tan^{-1}(2) \]