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The problem asks us to find an antiderivative of the function 

\[
\frac{1}{x\sqrt{x^2 - 16}}.
\]

Then, it requires evaluating the improper integral 

\[
\int_4^\infty \frac{dx}{x\sqrt{x^2 - 16}},
\]

by splitting it into two separate limits:

\[
\int_4^\infty \frac{dx}{x\sqrt{x^2 - 16}} = \lim_{t \to 4^+} \int_t^8 \frac{dx}{x\sqrt{x^2 - 16}} + \lim_{s \to \infty} \int_8^s \frac{dx}{x\sqrt{x^2 - 16}}.
\]

To find the antiderivative of \(\frac{1}{x\sqrt{x^2 - 16}}\), we can use a trigonometric substitution.

### Step 1: Substitution

Use the substitution \(x = 4\sec\theta\). Then:

- \(dx = 4\sec\theta\tan\theta \, d\theta\)
- \(\sqrt{x^2 - 16} = \sqrt{(4\sec\theta)^2 - 16} = 4\tan\theta\)

Substituting these into the integral:

\[
\int \frac{dx}{x\sqrt{x^2 - 16}} = \int \frac{4\sec\theta\tan\theta \, d\theta}{4\sec\theta \cdot 4\tan\theta} = \int \frac{d\theta}{4} = \frac{\theta}{4} + C.
\]

Since \(x = 4\sec\theta\), then \(\theta = \sec^{-1}\left(\frac{x}{4}\right)\).

### Step 2: Substituting back

The antiderivative is:

\[
\frac{\theta}{4} = \frac{\sec^{-1}\left(\frac{x}{4}\right)}{4} + C.
\]

### Step 3: Evaluating the Improper Integral

Evaluate the improper integral using the limits provided:

1. \(\lim_{t \to 4^+} \int_t^8 \frac{dx}{x\sqrt{x^2 - 16}}\).
2. \(\lim_{s \to \infty} \int_8^s \frac{dx}{x\sqrt{x^2 - 16}}\).

Would you like further assistance in computing these limits, or do you have any questions about this process?

### Relative Questions:
1. What is the purpose of trigonometric substitution in integration?
2. How do you evaluate an improper integral with infinite bounds?
3. Why is the function \(\sec^{-1}(x)\) used as the antiderivative?
4. What are the conditions for splitting an improper integral?
5. How do limits affect the convergence of an integral?

### Tip:
For improper integrals, always check the convergence by analyzing the behavior at the bounds before integrating.

Find an antiderivative for 1 / (x√(x^2 - 16)) and solve the improper integral.

Solve for the antiderivative of 1 / (x√(x^2 - 16)) using trigonometric substitution. Learn how to compute improper integrals with detailed steps and explanation of limits. Suitable for advanced high school or early college students.

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