Evaluate
`int_(0)^(oo)[(3)/(x^(2)+1)]dx`
where [.] denotes the greatest integer function.

Answer

To evaluate the integral

\[ I = \int_{0}^{\infty} \left\lfloor \frac{3}{x^2 + 1} \right\rfloor \, dx \]

where \(\left\lfloor . \right\rfloor\) denotes the greatest integer function, we need to analyze the function \(\frac{3}{x^2 + 1}\).

### Step 1: Determine the range of \(\frac{3}{x^2 + 1}\)

1. **Find the maximum value of \(\frac{3}{x^2 + 1}\)**:
- The function \(\frac{3}{x^2 + 1}\) is a decreasing function for \(x \geq 0\).
- When \(x = 0\), \(\frac{3}{0^2 + 1} = 3\).
- As \(x \to \infty\), \(\frac{3}{x^2 + 1} \to 0\).
- Therefore, the maximum value is \(3\) at \(x = 0\).

2. **Find the value of \(x\) where \(\frac{3}{x^2 + 1} = 2\)**:
- Set \(\frac{3}{x^2 + 1} = 2\):
\[ 3 = 2(x^2 + 1) \implies 3 = 2x^2 + 2 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}}. \]

3. **Find the value of \(x\) where \(\frac{3}{x^2 + 1} = 1\)**:
- Set \(\frac{3}{x^2 + 1} = 1\):
\[ 3 = x^2 + 1 \implies x^2 = 2 \implies x = \sqrt{2}. \]

### Step 2: Define intervals based on the greatest integer function

- For \(0 \leq x < \frac{1}{\sqrt{2}}\), \(\frac{3}{x^2 + 1} \geq 2\) and thus \(\left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 2\).
- For \(\frac{1}{\sqrt{2}} \leq x < \sqrt{2}\), \(\frac{3}{x^2 + 1} < 2\) and \(\geq 1\) so \(\left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 1\).
- For \(x \geq \sqrt{2}\), \(\frac{3}{x^2 + 1} < 1\) so \(\left\lfloor \frac{3}{x^2 + 1} \right\rfloor = 0\).

### Step 3: Break the integral into parts

Now we can break the integral into three parts:

\[ I = \int_{0}^{\frac{1}{\sqrt{2}}} 2 \, dx + \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\infty} 0 \, dx. \]

### Step 4: Evaluate each integral

1. **First integral**:
\[ \int_{0}^{\frac{1}{\sqrt{2}}} 2 \, dx = 2 \left[ x \right]_{0}^{\frac{1}{\sqrt{2}}} = 2 \left( \frac{1}{\sqrt{2}} - 0 \right) = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

2. **Second integral**:
\[ \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} 1 \, dx = \left[ x \right]_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} = \sqrt{2} - \frac{1}{\sqrt{2}} = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}. \]

3. **Third integral**:
\[ \int_{\sqrt{2}}^{\infty} 0 \, dx = 0. \]

### Step 5: Combine results

Now, we combine the results of the integrals:

\[ I = \sqrt{2} + \frac{1}{\sqrt{2}} + 0 = \sqrt{2} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{\frac{3}{\sqrt{2}}}. \]