# HHow do I integrate f(x)= x^2 e^(-5x) by parts?

Jan 28, 2015

Here you have to integrate by parts twice to "kill" your ${x}^{2}$.
$\int {x}^{2} {e}^{- 5 x} \mathrm{dx} = {x}^{2} \cdot {e}^{- 5 x} / \left(- 5\right) - \int {e}^{- 5 x} / \left(- 5\right) \cdot 2 x \mathrm{dx} =$
$= {x}^{2} \cdot {e}^{- 5 x} / \left(- 5\right) + \frac{1}{5} \int {e}^{- 5 x} \cdot 2 x \mathrm{dx} =$
$= {x}^{2} \cdot {e}^{- 5 x} / \left(- 5\right) + \frac{1}{5} \left[{e}^{- 5 x} / \left(- 5\right) \cdot 2 x - \int {e}^{- 5 x} / \left(- 5\right) \cdot 2 \mathrm{dx}\right] =$
$= {x}^{2} \cdot {e}^{- 5 x} / \left(- 5\right) + \frac{1}{5} \left[{e}^{- 5 x} / \left(- 5\right) \cdot 2 x + \frac{2}{5} \int {e}^{- 5 x} \mathrm{dx}\right] =$
$= {x}^{2} \cdot {e}^{- 5 x} / \left(- 5\right) + \frac{1}{5} {e}^{- 5 x} / \left(- 5\right) \cdot 2 x + \frac{1}{5} \cdot \frac{2}{5} {e}^{- 5 x} / \left(- 5\right) =$
$= {e}^{- 5 x} \left[- {x}^{2} / 5 - \frac{2 x}{25} - \frac{2}{125}\right] + c$