1. **State the problem:** We need to evaluate the integral $$\int (8x - 3) \ln x \, dx.$$\n\n2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du.$$\nWe choose parts to simplify the integral. Let:\n$$u = \ln x \implies du = \frac{1}{x} dx,$$\n$$dv = (8x - 3) dx \implies v = \int (8x - 3) dx = 4x^2 - 3x.$$\n\n3. **Apply integration by parts:**\n$$\int (8x - 3) \ln x \, dx = (4x^2 - 3x) \ln x - \int (4x^2 - 3x) \cdot \frac{1}{x} dx.$$\n\n4. **Simplify the integral:**\nInside the integral, simplify the integrand:\n$$ (4x^2 - 3x) \cdot \frac{1}{x} = 4x - 3.$$\nSo the integral becomes:\n$$ (4x^2 - 3x) \ln x - \int (4x - 3) dx.$$\n\n5. **Integrate the remaining integral:**\n$$\int (4x - 3) dx = 2x^2 - 3x + C.$$\n\n6. **Write the final answer:**\n$$\int (8x - 3) \ln x \, dx = (4x^2 - 3x) \ln x - (2x^2 - 3x) + C.$$\n\nThis is the evaluated integral.