Use the reduction formulas in a table of integrals to evaluate the following integrals. int p^2e^{-3p}dp

Question
Integrals
Use the reduction formulas in a table of integrals to evaluate the following integrals.
$$\int p^2e^{-3p}dp$$

2020-12-26
$$\text{Reduction formula for Exponential function is given as:}$$
$$\int x^ae^{bx}dx=[\frac{1}{b}x^ae^{bx}]-[\frac{a}{b}\int x^{a-1}e^{bc}]dx$$
$$\text{we have to evaluate:}$$
$$\int p^2e^{-3p}dp$$
$$\text{Using reduction formula, we will integrate it.}$$
$$\int p^2e^{-3p}dp=[\frac{1}{-3}p^2e^{-3p}]-[\frac{2}{-3}\int p^{2-1}\cdot e^{-3p}]dp$$
$$=\frac{1}{3}p^2e^{-3p}+\frac{2}{3}\int pe^{-3p}dp$$
$$\text{Now, we will integrate} \int pe^{-3p}dp\ \text{using integration by parts}$$
$$\int pe^{-3p}dp=p\int e^{-3p}dp-\int[\frac{d}{dx}(p)\int e^{-3p}dp]dp$$
$$=p\cdot\frac{e^{-3p}}{-3}-\int[1\cdot\frac{e^{-3p}}{-3}]dp$$
$$=-\frac{pe^{-3p}}{3}+\frac{1}{3}\int e^{-3p}dp$$
$$=-\frac{pe^{-3p}}{3}+\frac{1}{3}\cdot\frac{e^{-3p}}{-3}+c$$
$$=-\frac{pe^{-3p}}{3}-\frac{1}{9}e^{-3p}+c$$
$$\text{Putting above integration value in equation, we get }$$
$$\int p^2e^{-3p}dp=-\frac{1}{3}p^2e^{-3p}+\frac{2}{3}[-\frac{pe^{-3p}}{3}-\frac{1}{9}e^{-3p}]+c$$
$$=-\frac{1}{3}p^2e^{-3p}-\frac{2}{9}pe^{-3p}-\frac{2}{27}e^{-3p}+c$$
$$=-e^{-3p}(\frac{1}{3}p^2+\frac{2}{9}+\frac{2}{27})+c$$
$$\text{Hence, this is required integration.}$$

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