The value of `int_(- sqrt(3) )^( sqrt(3)) ((d)/(dx ) ( tan^(-1) (1/ x)) + x^(3) ) dx` is

Options

A. `pi//2`
B. `pi//4`
C. `1`
D. none of these

Answer

To solve the integral

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

we can break it down into two parts:

1. \(\int_{-\sqrt{3}}^{\sqrt{3}} \frac{d}{dx} \tan^{-1} \left( \frac{1}{x} \right) dx\)
2. \(\int_{-\sqrt{3}}^{\sqrt{3}} x^3 dx\)

### Step 1: Evaluate the first integral

Using the Fundamental Theorem of Calculus, we know that:

\[ \int \frac{d}{dx} f(x) \, dx = f(x) + C \]

Thus,

\[ \int_{-\sqrt{3}}^{\sqrt{3}} \frac{d}{dx} \tan^{-1} \left( \frac{1}{x} \right) dx = \tan^{-1} \left( \frac{1}{x} \right) \Big|_{-\sqrt{3}}^{\sqrt{3}}. \]

Calculating the limits:

\[ \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) - \tan^{-1} \left( \frac{1}{-\sqrt{3}} \right). \]

### Step 2: Calculate \(\tan^{-1} \left( \frac{1}{\sqrt{3}} \right)\)

We know that:

\[ \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \]

and

\[ \tan^{-1} \left( \frac{1}{-\sqrt{3}} \right) = -\tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = -\frac{\pi}{6}. \]

Thus,

\[ \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) - \tan^{-1} \left( \frac{1}{-\sqrt{3}} \right) = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}. \]

### Step 3: Evaluate the second integral

Now we evaluate the second integral:

\[ \int_{-\sqrt{3}}^{\sqrt{3}} x^3 \, dx. \]

Since \(x^3\) is an odd function (i.e., \(f(-x) = -f(x)\)), the integral of an odd function over a symmetric interval \([-a, a]\) is zero:

\[ \int_{-\sqrt{3}}^{\sqrt{3}} x^3 \, dx = 0. \]

### Step 4: Combine the results

Now we combine the results from both integrals:

\[ \int_{-\sqrt{3}}^{\sqrt{3}} \left( \frac{d}{dx} \tan^{-1} \left( \frac{1}{x} \right) + x^3 \right) dx = \frac{\pi}{3} + 0 = \frac{\pi}{3}. \]

### Final Answer

Thus, the value of the integral is

\[ \frac{\pi}{3}. \]