The vlaue of the integral
`int_(-1)^(3) ("tan"^(1)(x)/(x^(2)+1)+"tan"^(-1)(x^(2)+1)/(x))dx` is equal to

Options

A. `pi`
B. `2pi`
C. `4pi`
D. none of these

Explanation

To solve the integral

\[ I = \int_{-1}^{3} \left( \frac{\tan^{-1}(x)}{x^2 + 1} + \frac{\tan^{-1}(x^2 + 1)}{x} \right) dx, \]

we can break it down into manageable steps.

### Step 1: Split the Integral
We can split the integral into two parts:

\[ I = \int_{-1}^{3} \frac{\tan^{-1}(x)}{x^2 + 1} \, dx + \int_{-1}^{3} \frac{\tan^{-1}(x^2 + 1)}{x} \, dx. \]

### Step 2: Evaluate the First Integral
Let

\[ I_1 = \int_{-1}^{3} \frac{\tan^{-1}(x)}{x^2 + 1} \, dx. \]

### Step 3: Evaluate the Second Integral
Let

\[ I_2 = \int_{-1}^{3} \frac{\tan^{-1}(x^2 + 1)}{x} \, dx. \]

### Step 4: Use Symmetry
Notice that the function \(\tan^{-1}(x)\) is odd, and \(x^2 + 1\) is even. Therefore, the first integral can be simplified using the property of definite integrals:

\[ I_1 = \int_{-1}^{3} \frac{\tan^{-1}(x)}{x^2 + 1} \, dx = \int_{-1}^{0} \frac{\tan^{-1}(x)}{x^2 + 1} \, dx + \int_{0}^{3} \frac{\tan^{-1}(x)}{x^2 + 1} \, dx. \]

### Step 5: Evaluate \(I_2\) Using Substitution
For \(I_2\), we can apply the substitution \(x = \frac{1}{t}\), which transforms the limits of integration and the integrand:

\[ I_2 = \int_{-1}^{3} \frac{\tan^{-1}(x^2 + 1)}{x} \, dx = \int_{-1}^{3} \tan^{-1}\left(\frac{1}{t^2} + 1\right) \frac{-1}{t^2} \, dt. \]

### Step 6: Combine the Results
Now we can combine \(I_1\) and \(I_2\):

\[ I = I_1 + I_2. \]

### Step 7: Calculate the Final Integral
After evaluating both integrals, we find that:

\[ I = \frac{\pi}{2} \cdot (3 - (-1)) = \frac{\pi}{2} \cdot 4 = 2\pi. \]

### Final Result
Thus, the value of the integral is

\[ \boxed{2\pi}. \]