How do you integrate x3ex by integration by parts method?

1 Answer
Feb 2, 2017

Using integration by parts: =ex(x33x2+6x6)+C

Explanation:

The equation for integration by parts is this:

math.info

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=x3 so du=3x2dx

That leaves the rest of the integral to be dv

let dv=exdx so by taking the integral of both sides, v=ex

To recap, so far, we have

  1. u=x3
  2. du=3x2dx
  3. dv=exdx
  4. v=ex

soudv=uvvdu
=x3exex3x2dx

So lets go again!

  1. u=3x2
  2. du=6xdx
  3. dv=exdx
  4. v=ex

x3exex3x2dx
=x3ex(3x2exex6xdx)

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

  1. u=6x
  2. du=6dx
  3. dv=exdx
  4. v=ex

=x3ex(3x2exex6xdx)
=x3ex(3x2ex(6xexex6dx))

Simplify

=x3ex(3x2ex(6xex6ex))
=x3ex(3x2ex6xex+6ex)
=x3ex3x2ex+6xex6ex

Then you can take out the ex term if you'd like

=ex(x33x2+6x6)

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

=ex(x33x2+6x6)+C