evaluate the following integral. ∫6xe^2x dx

# Explanation: ## Step1: Apply integration - by - parts formula The integration - by - parts formula is $\int u\;dv=uv-\int v\;du$. Let $u = 6x$ and $dv=e^{2x}dx$. Then $du = 6dx$ and $v=\frac{1}{2}e^{2x}$. ## Step2: Substitute into the formula $\int 6xe^{2x}dx=6x\cdot\frac{1}{2}e^{2x}-\int\frac{1}{2}e^{2x}\cdot6dx$. Simplify to get $3xe^{2x}-3\int e^{2x}dx$. ## Step3: Integrate $e^{2x}$ We know that $\int e^{2x}dx=\frac{1}{2}e^{2x}+C$. So $3xe^{2x}-3\int e^{2x}dx=3xe^{2x}-3\cdot\frac{1}{2}e^{2x}+C$. # Answer: $3xe^{2x}-\frac{3}{2}e^{2x}+C$