To solve the given integral:

$\int_1^e \frac{\ln^2(x)}{x} \, dx$

We can use the substitution $u = \ln(x)$, which implies: $du = \frac{1}{x} dx \quad \text{and} \quad u \text{ ranges from } \ln(1) = 0 \text{ to } \ln(e) = 1.$

The integral then transforms as follows: $\int_1^e \frac{\ln^2(x)}{x} \, dx = \int_0^1 u^2 \, du.$

Solving the simplified integral:

$\int_0^1 u^2 \, du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}.$

Thus, the value of the integral is: $\boxed{\frac{1}{3}}$

This corresponds to option $D$.

Would you like a detailed explanation of substitution or the integration process?
Here are 5 related questions to expand your understanding:

1. How is substitution applied in integral calculus?
2. What is the geometric interpretation of the integral in this problem?
3. Can you solve $\int \frac{\ln(x)}{x} \, dx$ and compare?
4. What if the limits were different, such as $[1, e^2]$?
5. What is the connection between logarithmic functions and integration?

Tip: Always check the domain of functions like $\ln(x)$ before integrating to ensure the limits are valid!