Question

How do you integrate `int x^3e^x` by integration by parts method?

Answer 1

Using integration by parts: `=e^x ( x^3 - 3x^2 + 6x - 6)+C`

Explanation:

The equation for integration by parts is this:

I think mathematicians use similar letters on purpose to confuse people. Thats why they chose `u` and `v`, the worst possible letters.

Anyways, we have to pick our `u`.

let `u = x^3` so `du = 3x^2dx`

That leaves the rest of the integral to be `dv`

let `dv = e^xdx` so by taking the integral of both sides, `v = e^x`

To recap, so far, we have

1. `u = x^3`
2. `du = 3x^2dx`
3. `dv = e^xdx`
4. `v = e^x`

so`int u dv = uv - int v du`
`=x^3*e^x - inte^x*3x^2dx`

So lets go again!

1. `u = 3x^2`
2. `du = 6xdx`
3. `dv = e^xdx`
4. `v = e^x`

`x^3*e^x - inte^x*3x^2dx`
`= x^3*e^x - (3x^2*e^x - inte^x*6xdx)`

So lets go again (because calculus is so fun)!

1. `u = 6x`
2. `du = 6dx`
3. `dv = e^xdx`
4. `v = e^x`

`= x^3*e^x - (3x^2*e^x - inte^x*6xdx)`
`= x^3*e^x - (3x^2*e^x - (6x*e^x - int e^x*6dx))`

Simplify

`= x^3*e^x - (3x^2*e^x - (6x*e^x - 6e^x))`
`= x^3*e^x - (3x^2*e^x - 6x*e^x + 6e^x)`
`= x^3*e^x - 3x^2*e^x + 6x*e^x - 6e^x`

Then you can take out the `e^x` term if you'd like

`=e^x ( x^3 - 3x^2 + 6x - 6)`

Don't forget to add + C at the end, to signify that there is a group of equations what this integral could be.

`=e^x ( x^3 - 3x^2 + 6x - 6)+C`