evaluate the following integral. ∫ 5xe^(-3x) ...

# Explanation: ## Step1: Apply integration - by - parts formula The integration - by - parts formula is $\int u\;dv=uv-\int v\;du$. Let $u = 5x$ and $dv=e^{-3x}dx$. Then $du = 5dx$ and $v=-\frac{1}{3}e^{-3x}$. ## Step2: Substitute into the formula $\int 5xe^{-3x}dx=5x\left(-\frac{1}{3}e^{-3x}\right)-\int\left(-\frac{1}{3}e^{-3x}\right)\times5dx$. ## Step3: Simplify the first term and the new integral The first term is $-\frac{5}{3}xe^{-3x}$. The new integral is $\frac{5}{3}\int e^{-3x}dx$. ## Step4: Evaluate the remaining integral We know that $\int e^{-3x}dx=-\frac{1}{3}e^{-3x}+C$. So $\frac{5}{3}\int e^{-3x}dx=\frac{5}{3}\times\left(-\frac{1}{3}e^{-3x}\right)+C=-\frac{5}{9}e^{-3x}+C$. ## Step5: Combine the terms $\int 5xe^{-3x}dx=-\frac{5}{3}xe^{-3x}+\frac{5}{9}e^{-3x}+C=-\frac{5}{3}e^{-3x}\left(x+\frac{1}{3}\right)+C$. # Answer: $-\frac{5}{3}xe^{-3x}+\frac{5}{9}e^{-3x}+C$