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

Use the reduction formulas in a table of integrals to evaluate the following integrals.
$\int {p}^{2}{e}^{-3p}dp$
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grbavit
$\text{Reduction formula for Exponential function is given as:}$
$\int {x}^{a}{e}^{bx}dx=\left[\frac{1}{b}{x}^{a}{e}^{bx}\right]-\left[\frac{a}{b}\int {x}^{a-1}{e}^{bc}\right]dx$
$\text{we have to evaluate:}$
$\int {p}^{2}{e}^{-3p}dp$
$\text{Using reduction formula, we will integrate it.}$
$\int {p}^{2}{e}^{-3p}dp=\left[\frac{1}{-3}{p}^{2}{e}^{-3p}\right]-\left[\frac{2}{-3}\int {p}^{2-1}\cdot {e}^{-3p}\right]dp$
$=\frac{1}{3}{p}^{2}{e}^{-3p}+\frac{2}{3}\int p{e}^{-3p}dp$

$\int p{e}^{-3p}dp=p\int {e}^{-3p}dp-\int \left[\frac{d}{dx}\left(p\right)\int {e}^{-3p}dp\right]dp$
$=p\cdot \frac{{e}^{-3p}}{-3}-\int \left[1\cdot \frac{{e}^{-3p}}{-3}\right]dp$
$=-\frac{p{e}^{-3p}}{3}+\frac{1}{3}\int {e}^{-3p}dp$
$=-\frac{p{e}^{-3p}}{3}+\frac{1}{3}\cdot \frac{{e}^{-3p}}{-3}+c$
$=-\frac{p{e}^{-3p}}{3}-\frac{1}{9}{e}^{-3p}+c$

$\int {p}^{2}{e}^{-3p}dp=-\frac{1}{3}{p}^{2}{e}^{-3p}+\frac{2}{3}\left[-\frac{p{e}^{-3p}}{3}-\frac{1}{9}{e}^{-3p}\right]+c$
$=-\frac{1}{3}{p}^{2}{e}^{-3p}-\frac{2}{9}p{e}^{-3p}-\frac{2}{27}{e}^{-3p}+c$
$=-{e}^{-3p}\left(\frac{1}{3}{p}^{2}+\frac{2}{9}+\frac{2}{27}\right)+c$
$\text{Hence, this is required integration.}$