The integral to solve is:

$\int_e^\infty \frac{1}{x (\log x)^2} \, dx.$

Solution:

1. Substitution: Let $u = \log x$. Then, $du = \frac{1}{x} dx$.

   • When $x = e$, $u = 1$.
   • When $x \to \infty$, $u \to \infty$.

   The integral becomes: $\int_1^\infty \frac{1}{u^2} \, du.$

2. Evaluate the New Integral: The integral of $\frac{1}{u^2}$ is: $\int \frac{1}{u^2} \, du = -\frac{1}{u}.$

   So the definite integral is: $\left[-\frac{1}{u}\right]_1^\infty.$

3. Apply Limits: Evaluate the limits: $-\frac{1}{u} \Big|_1^\infty = \left(0 - (-1)\right) = 1.$

Final Answer:

$\int_e^\infty \frac{1}{x (\log x)^2} \, dx = 1.$

Let me know if you need further clarification!

Related Questions:

1. What happens if the lower limit is changed to another value greater than $e$?
2. How can substitution help simplify integrals like this?
3. What if the integrand was $\frac{1}{x (\log x)^3}$? How would the result change?
4. Can this integral be solved numerically without substitution?
5. What are the conditions for convergence of improper integrals like this one?

Tip:

For improper integrals, always check if the integral converges before solving. The behavior of the integrand as it approaches infinity or a singularity is key!