The value of the integral `int_(-4)^(4)[tan^(-1)((x)/(x^(4)+1))+tan^(-1)((x^(4)+1)/(x))]dx` equals

Options

A. `2pi`
B. `3pi`
C. `4pi`
D. `pi`

Answer

To solve the integral

\[ I = \int_{-4}^{4} \left[ \tan^{-1}\left(\frac{x}{x^4 + 1}\right) + \tan^{-1}\left(\frac{x^4 + 1}{x}\right) \right] dx, \]

we will use properties of inverse trigonometric functions.

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

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \quad \text{(for } x > 0\text{)}. \]

Thus, we can rewrite the integral as:

\[ I = \int_{-4}^{4} \left[ \tan^{-1}\left(\frac{x}{x^4 + 1}\right) + \tan^{-1}\left(\frac{x^4 + 1}{x}\right) \right] dx = \int_{-4}^{4} \frac{\pi}{2} dx. \]

### Step 2: Simplify the integral
Now, we can simplify the integral:

\[ I = \frac{\pi}{2} \int_{-4}^{4} dx. \]

### Step 3: Calculate the definite integral
The integral of \(dx\) from \(-4\) to \(4\) is:

\[ \int_{-4}^{4} dx = [x]_{-4}^{4} = 4 - (-4) = 8. \]

### Step 4: Substitute back into the equation
Now substituting this back into our expression for \(I\):

\[ I = \frac{\pi}{2} \cdot 8 = 4\pi. \]

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

\[ \boxed{4\pi}. \]