evaluate the following integral. ∫ 6x ln(3x) ...

# Explanation: ## Step1: Apply integration - by - parts formula The integration - by - parts formula is $\int u\;dv=uv-\int v\;du$. Let $u = \ln(3x)$ and $dv = 6x\;dx$. Then $du=\frac{1}{x}dx$ and $v=\int6x\;dx = 3x^{2}$. ## Step2: Substitute into the formula $\int6x\ln(3x)dx=3x^{2}\ln(3x)-\int3x^{2}\cdot\frac{1}{x}dx$. ## Step3: Simplify the second integral $\int3x^{2}\cdot\frac{1}{x}dx=\int3x\;dx=\frac{3}{2}x^{2}+C$. ## Step4: Write the final result $\int6x\ln(3x)dx = 3x^{2}\ln(3x)-\frac{3}{2}x^{2}+C$. # Answer: $3x^{2}\ln(3x)-\frac{3}{2}x^{2}+C$