How do you find the integral of  (x^2)(e^(-x))?

1 Answer
Sep 23, 2015

${x}^{2} {e}^{- x} - 2 x {e}^{- x} + 2 {e}^{- x} + c$

Explanation:

$\int {x}^{2} {e}^{- x} \mathrm{dx}$
if f(x)=x^2e^(-x) let u(x)=x^2 and v(x)=e^(-x)
using integration by parts
${x}^{2} {e}^{- x} \mathrm{dx} = {x}^{2} {e}^{- x} / - 1 - \int 2 x \cdot {e}^{- x} / - 1 = {x}^{2} {e}^{- x} + 2 \int x {e}^{- x} \mathrm{dx} = {x}^{2} {e}^{- x} + 2 \left[x {e}^{- x} / - 1 - {e}^{- x} / - 1\right] = {x}^{2} {e}^{- x} - 2 x {e}^{- x} + 2 {e}^{- x} + c$