The given integral is:

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

To solve this, let's proceed step by step:

Step 1: Substitution

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

Step 2: Adjust the Bounds

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

Thus, the integral becomes: $\int_{2}^{4} \frac{1}{u} \, du.$

Step 3: Solve the Simplified Integral

The integral of $\frac{1}{u}$ is: $\int \frac{1}{u} \, du = \ln |u| + C.$

Applying this to the bounds: $\int_{2}^{4} \frac{1}{u} \, du = \ln(4) - \ln(2).$

Step 4: Simplify the Result

Using the logarithmic property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$: $\ln(4) - \ln(2) = \ln\left(\frac{4}{2}\right) = \ln(2).$

Final Answer:

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

Would you like me to explain any specific part in more detail? Here are some related questions you might find interesting:

1. What is the significance of substitution in solving integrals?
2. How do logarithmic properties simplify integral evaluations?
3. Can this technique be applied to other forms of logarithmic integrals?
4. What happens if the bounds of the integral are swapped?
5. How does the change of variables affect definite integrals?

Tip: Always double-check your substitution limits after changing variables in definite integrals!