To determine whether the improper integral

$\int_{4}^{\infty} \frac{1}{x \sqrt{x^2 - 16}} \, dx$

converges or diverges, we need to analyze the behavior of the integrand, particularly as $x$ approaches the upper limit ($\infty$) and also ensure there are no issues at $x = 4$ (the lower limit).

Step 1: Analyze the Behavior at $x = 4$

At $x = 4$, we have the square root term $\sqrt{x^2 - 16}$. When $x = 4$, this term becomes $\sqrt{4^2 - 16} = \sqrt{0} = 0$. This suggests that the integrand might have a singularity at $x = 4$, making it potentially improper. We should check if the integrand approaches infinity as $x \to 4^+$.

At $x = 4$, the integrand behaves like:

$\frac{1}{x \sqrt{x^2 - 16}} \sim \frac{1}{x \cdot (x-4)} \quad \text{as} \quad x \to 4^+$

This gives a sense of how the function behaves near $x = 4$, and we'll need to check the limit as $x$ approaches 4 from the right.

Step 2: Check the Behavior as $x \to \infty$

We also need to check the behavior of the integrand as $x \to \infty$. To simplify:

• For large $x$, $\sqrt{x^2 - 16} \sim x$, so the integrand behaves like:

$\frac{1}{x \sqrt{x^2 - 16}} \sim \frac{1}{x \cdot x} = \frac{1}{x^2}$

We know that the integral of $1/x^2$ from 1 to infinity converges, so it is likely that the given integral converges as $x \to \infty$.

Step 3: Investigate the Improper Integral

Given that the integrand behaves as $\frac{1}{x \cdot (x-4)}$ near $x = 4$, we now perform the integration directly or consider whether this behavior leads to a divergent result at $x = 4$.

Step 4: Solve the Integral

Let's try to solve the integral.

Start with the substitution:

$x = 4 \sec(\theta), \quad dx = 4 \sec(\theta) \tan(\theta) \, d\theta$

Then, the integrand transforms as:

$\sqrt{x^2 - 16} = \sqrt{16(\sec^2(\theta) - 1)} = 4 \tan(\theta)$

Substitute into the integral:

$\int_{4}^{\infty} \frac{1}{x \sqrt{x^2 - 16}} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{4 \sec(\theta) \tan(\theta)}{4 \sec(\theta) \cdot 4 \tan(\theta)} \, d\theta$

Simplifying:

$= \int_{0}^{\frac{\pi}{2}} \frac{1}{4} \, d\theta = \frac{\pi}{8}$

Conclusion:

The improper integral converges and its value is:

$\boxed{\frac{\pi}{8}}$

Would you like further details or have any other questions?

Here are some related questions to explore:

1. How can substitution methods help with other integrals involving square roots?
2. What would happen if the integrand had a different form, like $\sqrt{x^2 - 16}$ in the numerator?
3. Can the substitution $x = 4 \sec(\theta)$ be used for other types of integrals involving square roots?
4. What does the convergence of this integral tell us about the behavior of functions at singular points?
5. How does the comparison test help in analyzing improper integrals?

Tip: When facing integrals involving square roots and rational functions, consider substitution or trigonometric identities to simplify the expression and make integration easier.