This integral is:

$\int_{e^2}^{e^4} \frac{dx}{x \ln x}.$

Step-by-Step Solution:

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

   This transforms the integral into: $\int \frac{1}{u} \, du.$

2. Adjusting the Limits:

   • When $x = e^2$, $u = \ln(e^2) = 2$.
   • When $x = e^4$, $u = \ln(e^4) = 4$.

   So the limits of the integral are updated to $u \in [2, 4]$.

3. Simplified Integral: The integral becomes: $\int_{2}^{4} \frac{1}{u} \, du.$

4. Evaluate: The integral of $\frac{1}{u}$ is $\ln |u|$, so: $\int_{2}^{4} \frac{1}{u} \, du = \left[ \ln u \right]_{2}^{4}.$

5. Compute the Result: Substitute the limits: $\ln(4) - \ln(2) = \ln\left(\frac{4}{2}\right) = \ln(2).$

Final Answer:

$\int_{e^2}^{e^4} \frac{dx}{x \ln x} = \ln(2).$

Let me know if you'd like further clarification!

Related Questions:

1. How does substitution simplify this problem?
2. What are other applications of logarithmic substitution in integration?
3. Can this integral be solved without substitution? If so, how?
4. Why is the logarithmic property $\ln(a) - \ln(b) = \ln(a/b)$ crucial in this context?
5. What would happen if the limits were reversed (e.g., from $e^4$ to $e^2$)?

Tip: Always verify the new limits when using substitution, as they must align with the transformed variable!