Evaluate: `int_(-1)^3(tan^(-1)(x/(x^2+1))+tan^(-1)((x^2+1)/x))dx`

Explanation

To evaluate the integral

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

we can use a property of the inverse tangent function.

### Step 1: Use the property of inverse tangent
We know that:

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \]

for \(x > 0\). In our case, we can express the second term as follows:

\[ \tan^{-1}\left(\frac{x^2 + 1}{x}\right) = \tan^{-1}\left(\frac{1}{\frac{x}{x^2 + 1}}\right). \]

### Step 2: Rewrite the integral
Thus, we can rewrite the integral as:

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

because the two terms combine to give \(\frac{\pi}{2}\).

### Step 3: Simplify the integral
Now we can simplify the integral:

\[ I = \int_{-1}^{3} \frac{\pi}{2} \, dx. \]

### Step 4: Evaluate the integral
Since \(\frac{\pi}{2}\) is a constant, we can factor it out:

\[ I = \frac{\pi}{2} \int_{-1}^{3} dx. \]

Now, we calculate the integral:

\[ \int_{-1}^{3} dx = [x]_{-1}^{3} = 3 - (-1) = 3 + 1 = 4. \]

### Step 5: Final calculation
Substituting back, we get:

\[ I = \frac{\pi}{2} \cdot 4 = 2\pi. \]

Thus, the final answer is:

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