Integral X Ln3X Fe1Bc6

1. **State the problem:** We want to find the integral $$\int x \ln(3x) \, dx.$$\n\n2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du.$$\n\n3. **Choose parts:** Let $$u = \ln(3x)$$ and $$dv = x \, dx.$$\n\n4. **Compute derivatives and integrals:** $$du = \frac{1}{3x} \cdot 3 \, dx = \frac{1}{x} \, dx,$$ simplifying the derivative of $$\ln(3x)$$ using chain rule.\n$$v = \int x \, dx = \frac{x^2}{2}.$$\n\n5. **Apply integration by parts:** $$\int x \ln(3x) \, dx = \frac{x^2}{2} \ln(3x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln(3x) - \frac{1}{2} \int x \, dx.$$\n\n6. **Simplify the remaining integral:** $$\int x \, dx = \frac{x^2}{2}.$$\n\n7. **Substitute back:** $$\int x \ln(3x) \, dx = \frac{x^2}{2} \ln(3x) - \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{2} \ln(3x) - \frac{x^2}{4} + C,$$ where $$C$$ is the constant of integration.\n\n**Final answer:** $$\boxed{\int x \ln(3x) \, dx = \frac{x^2}{2} \ln(3x) - \frac{x^2}{4} + C}.$$