How do you integrate (x-x^2)e^x?

Yep, you’ll need integration by parts! Two rounds of it. Lucky you :P

Start by choosing one term to be u and one term to be dv. You’ll have to take the derivative of u and integrate dv. The important thing is to always choose u to be something that gets simpler when you take the derivative. e^x remains e^x when you derive, so don’t choose that.

u = (x-x^2), dv = e^x
du = 1–2x, v = e^x

Now your formula is:

integral(udv) = uv – integral(vdu)
So integral((x-x^2)e^x) = (x-x^2)e^x – integral((1–2x)e^x)
But in order to solve that second integral, (1–2x)e^x, you have to use integration by parts again!

Choose u = 1–2x, dv = e^x
du = -2, v = e^x

integral((1–2x)e^x) = (1–2x)e^x – integral(-2e^x)

Okay, NOW the integral on the righthand side is easily solvable! In fact its integral is itself: -2e^x.

Plug THAT in to the first result from integration by parts:

integral((x-x^2)e^x) = (x-x^2)e^x – integral((1–2x)e^x)
= (x-x^2)e^x – ((1–2x)e^x – integral(-2e^x))
= (x-x^2)e^x – ((1–2x)e^x – (-2e^x))

Now just distribute your negative signs:
= (x-x^2)e^x – (1–2x)e^x – 2e^x
You could leave it there, or clean it up a bit by distributing your terms and combining like terms:
xe^x – x^2e^x – e^x + 2xe^x – 2e^x
-x^2e^x + 3xe^x -3e^x

Done!

relevant: http://xkcd.com/1201/

Yep….. This is why I failed algebra.
I now have a migraine. lol

This is calculus!
Sorry for the headache!

lol Sorry! It’s all Greek to me!!

Big kudo’s to the people that understand it!!
I have to ask my 18 year old son for help with…... Fractions.
Math makes me cry. lol

I used to know this stuff but over the years in seems to have gone into one neuron and out through another!