`int_(-1)^(1)(d)/(dx)[tan^(-1)((1)/(x))]dx=`

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `-(pi)/(2)`
D. `-(pi)/(4)`

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Differentiate the function inside the integral

We need to find the derivative of \( \tan^{-1} \left( \frac{1}{x} \right) \). Using the chain rule, we have:

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

Calculating \( \frac{d}{dx} \left( \frac{1}{x} \right) \):

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

Now substituting this back, we get:

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

### Step 2: Simplify the expression

The term \( 1 + \frac{1}{x^2} \) can be rewritten as:

\[ 1 + \frac{1}{x^2} = \frac{x^2 + 1}{x^2}. \]

Thus, we have:

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

### Step 3: Set up the integral

Now we can substitute this back into our integral:

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

### Step 4: Evaluate the integral

This simplifies to:

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

The integral \( \int \frac{1}{x^2 + 1} \, dx \) is known to be \( \tan^{-1}(x) \). Thus, we evaluate:

\[ I = -\left[ \tan^{-1}(x) \right]_{-1}^{1}. \]

Calculating the limits:

\[ \tan^{-1}(1) = \frac{\pi}{4}, \quad \tan^{-1}(-1) = -\frac{\pi}{4}. \]

So,

\[ I = -\left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = -\left( \frac{\pi}{4} + \frac{\pi}{4} \right) = -\frac{\pi}{2}. \]

### Final Result

Thus, the value of the integral is:

\[ I = -\frac{\pi}{2}. \]

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