1. The problem is to evaluate the integral using integration by parts formula: $$\int u\,dv = uv - \int v\,du$$ 2. Given: $$u = \ln x \implies du = \frac{1}{x} dx$$ $$dv = dx \implies v = x$$ 3. Substitute into the formula: $$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx$$ 4. Simplify the integral: $$\int x \cdot \frac{1}{x} dx = \int \cancel{x} \cdot \frac{1}{\cancel{x}} dx = \int 1 \, dx$$ 5. Evaluate the remaining integral: $$\int 1 \, dx = x + C$$ 6. Therefore, the solution is: $$\int \ln x \, dx = x \ln x - x + C$$ where $C$ is the constant of integration.