evaluate the following integral. ∫9xe^(-2x) dx

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = 9x$ and $dv=e^{-2x}dx$. Then $du = 9dx$ and $v=-\frac{1}{2}e^{-2x}$.

Step2: Substitute into the formula

$\int 9xe^{-2x}dx=9x\left(-\frac{1}{2}e^{-2x}\right)-\int\left(-\frac{1}{2}e^{-2x}\right)\times9dx$.

Step3: Simplify the first term and the new integral

The first term is $-\frac{9}{2}xe^{-2x}$. The new integral is $\frac{9}{2}\int e^{-2x}dx$.

Step4: Evaluate the remaining integral

We know that $\int e^{-2x}dx=-\frac{1}{2}e^{-2x}+C$. So $\frac{9}{2}\int e^{-2x}dx=\frac{9}{2}\times\left(-\frac{1}{2}e^{-2x}\right)+C=-\frac{9}{4}e^{-2x}+C$.

Step5: Combine the terms

$\int 9xe^{-2x}dx=-\frac{9}{2}xe^{-2x}+\frac{9}{4}e^{-2x}+C=-\frac{9}{2}e^{-2x}\left(x-\frac{1}{2}\right)+C$.

Answer:

$-\frac{9}{2}xe^{-2x}+\frac{9}{4}e^{-2x}+C$