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

1 Answer
Feb 2, 2017

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