evaluate the following integral. ∫ 4xe^8x dx

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Let $u = 4x$ and $dv=e^{8x}dx$. Then $du = 4dx$ and $v=\frac{1}{8}e^{8x}$.

Step2: Substitute into the formula

$\int 4xe^{8x}dx=4x\cdot\frac{1}{8}e^{8x}-\int\frac{1}{8}e^{8x}\cdot4dx$. Simplify to get $\frac{1}{2}xe^{8x}-\frac{1}{2}\int e^{8x}dx$.

Step3: Integrate the remaining integral

We know that $\int e^{ax}dx=\frac{1}{a}e^{ax}+C$ (where $a = 8$ here). So $\int e^{8x}dx=\frac{1}{8}e^{8x}+C$. Then $\frac{1}{2}xe^{8x}-\frac{1}{2}\int e^{8x}dx=\frac{1}{2}xe^{8x}-\frac{1}{2}\cdot\frac{1}{8}e^{8x}+C$.

Step4: Simplify the result

$\frac{1}{2}xe^{8x}-\frac{1}{16}e^{8x}+C=\frac{e^{8x}(8x - 1)}{16}+C$.

Answer:

$\frac{e^{8x}(8x - 1)}{16}+C$