Question

How do you integrate `int x^2 e^(-x) dx` using integration by parts?

Answer 1

`I = -e^(-x)(x^2 + 2x + 2) + c`

Explanation:

`I = intx^2e^(-x)dx`

Say `u = x^2` so `du = 2x` and `dv = e^(-x)` so `v = -e^(-x)`

`I = -x^2e^(-x) + 2intxe^(-x)dx`

Say `u = x` so `du = 1` and `dv = e^(-x)` so `v = -e^(-x)`

`I = -x^2e^(-x) + 2(-xe^(-x) + inte^(-x)dx)`
`I = -x^2e^(-x) -2xe^(-x) + 2inte^(-x)dx`
`I = -x^2e^(-x)-2xe^(-x)-2e^(-x)+c`
`I = -e^(-x)(x^2 + 2x + 2) + c`