Use integral tables to evaluate the integral. \( \int \frac{d x}{x \sqrt{33+x^{2}}} \)

To evaluate the integral \( \int \frac{d x}{x \sqrt{33+x^{2}}} \), we can utilize a substitution. Let's set \( x = \sqrt{33} \tan(\theta) \). This implies that \( dx = \sqrt{33} \sec^{2}(\theta) d\theta \). Now the integral transforms as follows: 1. Substitute \( x \) into the integral: \[ \sqrt{33+x^{2}} = \sqrt{33 + 33\tan^{2}(\theta)} = \sqrt{33(1+\tan^{2}(\theta))} = \sqrt{33} \sec(\theta) \] 2. Now substitute \( dx \) and the \( \sqrt{33+x^{2}} \) into the integral: \[ \int \frac{\sqrt{33} \sec^{2}(\theta) d\theta}{\sqrt{33} \tan(\theta) \cdot \sqrt{33} \sec(\theta)} = \int \frac{\sec^{2}(\theta) d\theta}{\tan(\theta) \sec(\theta)} \] Simplifying this gives: \[ = \int \frac{\sec(\theta)}{\tan(\theta)} d\theta = \int \frac{1}{\sin(\theta)} d\theta \] 3. The integral of \( \frac{1}{\sin(\theta)} \) is: \[ = \ln | \csc(\theta) - \cot(\theta) | + C \] 4. Reverting back to \( x \) using \( \tan(\theta) = \frac{x}{\sqrt{33}} \): \[ \csc(\theta) = \frac{\sqrt{33}}{x}, \quad \text{and} \quad \cot(\theta) = \frac{\sqrt{33}}{x} \tan(\theta) = \frac{\sqrt{33}}{x} \cdot \frac{x}{\sqrt{33}} = 1 \] 5. Hence, the integral evaluates to: \[ = \ln \left| \frac{\sqrt{33}}{x} - 1 \right| + C \] So, the final result of the integral \( \int \frac{dx}{x \sqrt{33+x^{2}}} \) is: \[ \ln \left| \frac{\sqrt{33}}{x} - 1 \right| + C \]