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The given problem asks to evaluate the integral:

\[
\int \frac{dx}{x\sqrt{16x^2 - 2}}
\]

### Solution Approach:

This is a standard integral that can be solved using a trigonometric substitution. Let’s solve this step by step:

1. **Rewrite the integral**: The expression inside the square root is of the form \( 16x^2 - 2 \), which suggests a trigonometric substitution.
   
   Factor out the 16:
   \[
   16x^2 - 2 = 2(8x^2 - 1)
   \]

   Now, we will use a substitution based on the form of \( x^2 - a^2 \).

2. **Substitute**: Let \( x = \frac{1}{2\sqrt{2}} \sec(\theta) \). This substitution helps simplify the integral involving \( \sec^2(\theta) \) and the square root terms.

3. **Simplify**: Carry out the substitution and simplify the resulting trigonometric integral.

4. **Evaluate**: Solve the simplified integral and back-substitute to find the solution in terms of \( x \).

After evaluating, the correct answer matches option **D**:

\[
\boxed{D. \frac{\sqrt{2}}{2} \sec^{-1}\left|2\sqrt{2}x\right| + C}
\]

Would you like a more detailed walkthrough of the trigonometric substitution and solution steps?

### Relative Questions:

1. How do you identify appropriate trigonometric substitutions for integrals?
2. What is the purpose of factoring constants like 16 from the square root before substituting?
3. How does the secant function help in integrals involving square roots of quadratic terms?
4. What is the general strategy for solving integrals of the form \( \frac{1}{x \sqrt{ax^2 + b}} \)?
5. Why is it important to include the absolute value in inverse trigonometric function results?

### Tip:

When working with integrals that contain square roots of quadratic expressions, always consider trigonometric substitutions like \( x = a \sec(\theta) \) or \( x = a \sin(\theta) \) depending on the form of the expression.

Evaluate the integral: ∫ (dx) / (x√(16x² - 2))

Learn how to evaluate the integral ∫ (dx) / (x√(16x² - 2)) using trigonometric substitution techniques. This step-by-step solution involves secant function substitution and inverse trigonometric functions. Suitable for advanced high school and undergraduate students studying calculus.

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