# How do you integrate int x^2e^(-3x) by integration by parts method?

Jul 31, 2016

$= - {e}^{- 3 x} \left(\frac{1}{3} {x}^{2} + \frac{2}{9} x + \frac{2}{27}\right) + C$

#### Explanation:

$\int {x}^{2} {e}^{- 3 x} \setminus \mathrm{dx}$

$= \int {x}^{2} {\left(- \frac{1}{3} {e}^{- 3 x}\right)}^{p} r i m e \setminus \mathrm{dx}$

which by IBP, ie: $\int u v ' = u v - \int u ' v$

$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} - \int {\left({x}^{2}\right)}^{p} r i m e \left(- \frac{1}{3} {e}^{- 3 x}\right) \setminus \mathrm{dx}$

$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} + \frac{1}{3} \int 2 x {e}^{- 3 x} \setminus \mathrm{dx}$

$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} + \frac{2}{3} \int x {\left(- \frac{1}{3} {e}^{- 3 x}\right)}^{p} r i m e \setminus \mathrm{dx}$

which by IBP again
$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} + \frac{2}{3} \left(- \frac{1}{3} x {e}^{- 3 x} - \int {\left(x\right)}^{p} r i m e \left(- \frac{1}{3} {e}^{- 3 x}\right) \setminus \mathrm{dx}\right)$

$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} - \frac{2}{9} x {e}^{- 3 x} + \frac{2}{9} \int {e}^{- 3 x} \setminus \mathrm{dx}$

$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} - \frac{2}{9} x {e}^{- 3 x} + \frac{2}{9} \left(- \frac{1}{3} {e}^{- 3 x}\right) + C$

$= - \frac{1}{3} {x}^{2} {e}^{- 3 x} - \frac{2}{9} x {e}^{- 3 x} - \frac{2}{27} {e}^{- 3 x} + C$

$= - {e}^{- 3 x} \left(\frac{1}{3} {x}^{2} + \frac{2}{9} x + \frac{2}{27}\right) + C$