What is `int_(-1)^(1) {d/(dx) (tan^(-1)'1/x)} dx` equal to ?

Options

A. `0`
B. `-pi/4`
C. `-pi/2`
D. `pi/2`

Explanation

To solve the integral \( \int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \left( \frac{1}{x} \right) \right) dx \), we can follow these steps:

### Step 1: Recognize the Integral of a Derivative
We start with the integral of a derivative:
\[ \int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \left( \frac{1}{x} \right) \right) dx \]

### Step 2: Apply the Fundamental Theorem of Calculus
By the Fundamental Theorem of Calculus, we know that:
\[ \int_{a}^{b} f'(x) \, dx = f(b) - f(a) \]
Thus, we can write:
\[ \int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \left( \frac{1}{x} \right) \right) dx = \tan^{-1} \left( \frac{1}{1} \right) - \tan^{-1} \left( \frac{1}{-1} \right) \]

### Step 3: Evaluate the Arctangent Function
Now we evaluate the arctangent at the bounds:
\[ \tan^{-1}(1) = \frac{\pi}{4} \quad \text{and} \quad \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Step 4: Substitute the Values
Substituting these values into our expression gives:
\[ \tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Final Result
Thus, we find that:
\[ \int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \left( \frac{1}{x} \right) \right) dx = \frac{\pi}{2} \]