Question

How do you integrate `int x^2 e^(4-x) dx` using integration by parts?

Answer 1

`-x^2e^(4-x)-2xe^(4-x)+2e^(4-x)+const.`

Explanation:

For integration by parts it's used this scheme:

`int udv=uv-int vdu`

Making
`dv=e^(4-x)dx` => `v=-e^(4-x)`
`u=x^2` => `du=2xdx`
Then the original expression becomes
`=x^2(-e^(4-x))-int -e^(4-x)2xdx=-x^2e^(4-x)+2int xe^(4-x)dx`

Applying integration by parts to `int xe^(4-x)dx`
Making
`dv=e^(4-x)dx` => `v=-e^(4-x)`
`u=x` => `du=dx`
We get
`=x(-e^(4-x))-int e^(4-x)dx=-xe^(4-x)+e^(4-x)`

Using the partial results in the main expression, we get
`=-x^2e^(4-x)+2(-xe^(4-x)+e^(4-x))`
`=-x^2e^(4-x)-2xe^(4-x)+2e^(4-x)+const.`