To evaluate the integral $\int_{1}^{\infty} \frac{\ln(x)}{x} \, dx$, we'll go through the following steps:

1. Integral Setup: $I = \int_{1}^{\infty} \frac{\ln(x)}{x} \, dx$

2. Substitution: Let $u = \ln(x)$, so $du = \frac{1}{x} \, dx$.

   When $x = 1$, $u = \ln(1) = 0$.

   When $x \to \infty$, $u \to \infty$.

   The integral now becomes: $I = \int_{0}^{\infty} u \, du$

3. Evaluate the Integral: The integral of $u$ with respect to $u$ is straightforward: $I = \left. \frac{u^2}{2} \right|_{0}^{\infty}$

   Evaluating this: $I = \lim_{u \to \infty} \frac{u^2}{2} - \frac{0^2}{2} = \infty$

The integral diverges, meaning it does not converge to a finite value. Therefore:

$\int_{1}^{\infty} \frac{\ln(x)}{x} \, dx \text{ diverges.}$

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. What is an example of a similar integral that converges?
2. How can you determine whether an improper integral converges?
3. What is the integral of $\frac{1}{x} \ln(x)$ over a finite interval, such as $[1, e]$?
4. How does substitution affect the bounds of integration?
5. Can we use any other techniques, like integration by parts, to solve this integral?

Tip: When dealing with improper integrals, always check for divergence by evaluating limits, especially as the variable approaches infinity or a point of discontinuity.