Question

compute the given integral.
$int 6x^{3}e^{3x}dx=$
$+c$

Explanation:

Step1: Apply integration - by - parts formula $\int u\;dv=uv-\int v\;du$

Let $u = 6x^{3}$ and $dv=e^{3x}dx$. Then $du = 18x^{2}dx$ and $v=\frac{1}{3}e^{3x}$.
So, $\int 6x^{3}e^{3x}dx = 6x^{3}\cdot\frac{1}{3}e^{3x}-\int\frac{1}{3}e^{3x}\cdot18x^{2}dx=2x^{3}e^{3x}- 6\int x^{2}e^{3x}dx$.

Step2: Apply integration - by - parts again on $\int x^{2}e^{3x}dx$

Let $u = x^{2}$ and $dv=e^{3x}dx$. Then $du = 2xdx$ and $v=\frac{1}{3}e^{3x}$.
$\int x^{2}e^{3x}dx=x^{2}\cdot\frac{1}{3}e^{3x}-\int\frac{1}{3}e^{3x}\cdot2xdx=\frac{1}{3}x^{2}e^{3x}-\frac{2}{3}\int xe^{3x}dx$.

Step3: Apply integration - by - parts on $\int xe^{3x}dx$

Let $u = x$ and $dv=e^{3x}dx$. Then $du = dx$ and $v=\frac{1}{3}e^{3x}$.
$\int xe^{3x}dx=x\cdot\frac{1}{3}e^{3x}-\int\frac{1}{3}e^{3x}dx=\frac{1}{3}xe^{3x}-\frac{1}{9}e^{3x}+C_1$.

Step4: Substitute back

$\int x^{2}e^{3x}dx=\frac{1}{3}x^{2}e^{3x}-\frac{2}{3}(\frac{1}{3}xe^{3x}-\frac{1}{9}e^{3x})+C_2=\frac{1}{3}x^{2}e^{3x}-\frac{2}{9}xe^{3x}+\frac{2}{27}e^{3x}+C_2$.
$\int 6x^{3}e^{3x}dx=2x^{3}e^{3x}-6(\frac{1}{3}x^{2}e^{3x}-\frac{2}{9}xe^{3x}+\frac{2}{27}e^{3x})+C$
$=2x^{3}e^{3x}- 2x^{2}e^{3x}+\frac{4}{3}xe^{3x}-\frac{4}{9}e^{3x}+C$.

Answer:

$2x^{3}e^{3x}-2x^{2}e^{3x}+\frac{4}{3}xe^{3x}-\frac{4}{9}e^{3x}+C$