1. The problem is to evaluate the definite integral $$\int_{e^2}^{e^3} \frac{1}{x} \, dx$$. 2. Recall the integral formula for $$\int \frac{1}{x} \, dx = \ln|x| + C$$, where $$\ln$$ is the natural logarithm. 3. Applying the Fundamental Theorem of Calculus, we evaluate: $$\int_{e^2}^{e^3} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{e^2}^{e^3} = \ln(e^3) - \ln(e^2)$$. 4. Simplify the logarithms using the property $$\ln(a^b) = b \ln(a)$$: $$\ln(e^3) - \ln(e^2) = 3 \ln(e) - 2 \ln(e)$$. 5. Since $$\ln(e) = 1$$, this becomes: $$3 \times 1 - 2 \times 1 = 3 - 2 = 1$$. 6. Therefore, the value of the integral is $$1$$. Final answer: 1