`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 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 intervals for the greatest integer function
The greatest integer function \([x]\) takes different constant values in different intervals:
- For \( x \in [-1, 0) \), \([x] = -1\)
- For \( x \in [0, 1) \), \([x] = 0\)
- For \( x \in [1, 2) \), \([x] = 1\)

### Step 2: Break the integral into segments
We can break the integral \( I \) into three parts:
\[ I = \int_{-1}^{0} \frac{[x]}{1+x^2} \, dx + \int_{0}^{1} \frac{[x]}{1+x^2} \, dx + \int_{1}^{2} \frac{[x]}{1+x^2} \, dx \]

### Step 3: Evaluate each segment

#### Segment 1: \( \int_{-1}^{0} \frac{[x]}{1+x^2} \, dx \)
For \( x \in [-1, 0) \), \([x] = -1\):
\[ \int_{-1}^{0} \frac{-1}{1+x^2} \, dx = -\int_{-1}^{0} \frac{1}{1+x^2} \, dx \]
The integral of \(\frac{1}{1+x^2}\) is \(\tan^{-1}(x)\):
\[ -\left[ \tan^{-1}(x) \right]_{-1}^{0} = -\left( \tan^{-1}(0) - \tan^{-1}(-1) \right) = -\left( 0 - \left(-\frac{\pi}{4}\right) \right) = -\left( \frac{\pi}{4} \right) = -\frac{\pi}{4} \]

#### Segment 2: \( \int_{0}^{1} \frac{[x]}{1+x^2} \, dx \)
For \( x \in [0, 1) \), \([x] = 0\):
\[ \int_{0}^{1} \frac{0}{1+x^2} \, dx = 0 \]

#### Segment 3: \( \int_{1}^{2} \frac{[x]}{1+x^2} \, dx \)
For \( x \in [1, 2) \), \([x] = 1\):
\[ \int_{1}^{2} \frac{1}{1+x^2} \, dx = \left[ \tan^{-1}(x) \right]_{1}^{2} = \tan^{-1}(2) - \tan^{-1}(1) = \tan^{-1}(2) - \frac{\pi}{4} \]

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

### Final Result
Thus, the value of the integral is:
\[ I = \tan^{-1}(2) - \frac{\pi}{2} \]