The value of `int_(-1)^(1) (d)/(dx) ("tan"^(1)(1)/(x))dx` is

Options

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

Answer

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

### Step 1: Understanding the Integral
The integral involves the derivative of the function \( \tan^{-1} \left( \frac{1}{x} \right) \). According to the Fundamental Theorem of Calculus, the integral of a derivative over an interval gives us the difference of the values of the function at the endpoints of the interval.

### Step 2: Apply the Fundamental Theorem of Calculus
We can express the integral as:
\[ I = \left[ \tan^{-1} \left( \frac{1}{x} \right) \right]_{-1}^{1} \]

### Step 3: Evaluate the Function at the Limits
Now we need to evaluate \( \tan^{-1} \left( \frac{1}{x} \right) \) at the limits \( x = 1 \) and \( x = -1 \).

1. For \( x = 1 \):
\[ \tan^{-1} \left( \frac{1}{1} \right) = \tan^{-1}(1) = \frac{\pi}{4} \]

2. For \( x = -1 \):
\[ \tan^{-1} \left( \frac{1}{-1} \right) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

### Step 4: Calculate the Difference
Now we substitute these values back into the expression for \( I \):
\[ I = \tan^{-1} \left( \frac{1}{1} \right) - \tan^{-1} \left( \frac{1}{-1} \right) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \]
\[ I = \frac{\pi}{4} + \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{2}} \]