The given expression is an integral of the form:

$\int_{1}^{e^2} \frac{1}{x} \, dx$

This is a standard integral. The integral of $\frac{1}{x}$ with respect to $x$ is the natural logarithm of $x$:

$\int \frac{1}{x} \, dx = \ln |x| + C$

For a definite integral from $x = 1$ to $x = e^2$, we calculate:

$\int_{1}^{e^2} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{1}^{e^2}$

Now, substitute the limits into the expression:

$\left[ \ln |x| \right]_{1}^{e^2} = \ln |e^2| - \ln |1|$

Simplifying further:

$= \ln (e^2) - \ln (1)$

Since $\ln (e^2) = 2$ and $\ln (1) = 0$:

$= 2 - 0 = 2$

Thus, the value of the integral is $2$.

Would you like to ask any more questions or need further details?

Here are some related questions you might find useful:

1. How do you solve other types of definite integrals?
2. What is the significance of the natural logarithm in calculus?
3. Can you explain the process of integration by parts?
4. How does the fundamental theorem of calculus apply to definite integrals?
5. What are some common mistakes to avoid when evaluating definite integrals?

Tip: When solving definite integrals, always remember to evaluate the integral function at both the upper and lower limits to get the correct result.