Let's break down the expression you've provided:

$\ln(x) \cdot x + \int x \cdot \frac{1}{x} \, dx$

Step 1: Simplify the integral

The second term is an integral:

$\int x \cdot \frac{1}{x} \, dx$

Since $x \cdot \frac{1}{x} = 1$, this simplifies to:

$\int 1 \, dx = x + C$

where $C$ is the constant of integration.

Step 2: Combine the two terms

Now, the expression becomes:

$\ln(x) \cdot x + x + C$

This is the simplified result.

Final Answer:

The expression simplifies to:

$x \ln(x) + x + C$

Would you like more details on any part of this solution, or have any other questions? Here's a list of further related questions:

1. How do logarithmic and exponential functions behave in integration?
2. What are some other common integration techniques involving logarithms?
3. How do you handle indefinite integrals like this with constants of integration?
4. Can you explain the properties of logarithmic differentiation?
5. What are common rules for simplifying integrals of rational functions?

Tip: When integrating expressions involving $\frac{1}{x}$, remember that $\int \frac{1}{x} \, dx = \ln|x| + C$.