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

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:

$-\frac{5}{3}xe^{-3x}+\frac{5}{9}e^{-3x}+C$