Question

Integration help please..

Answer 1

\[-xe^{\frac{-x}{2}}\]
How do I integrate that?

Answer 2

By parts

Answer 3

Okay, so then I choose u = -x, dv = e^(-x/2)
and du = -1, v = int(e^(-x/2)) dx?

Answer 4

\[\huge -\int\limits xe^{-x/2}dx \\ \\ \huge u=x~~~~~dv=e^{-x/2}dx \\ \\ \huge du=dx~~~~~v=-2 e^{-\frac{x}{2}} \\ \\ \huge -[-2xe^{\frac{-x}{2}}- \int\limits (-2)e^{\frac{-x}{2}}dx]\]
Then you just integrate e^(-x/2) 1 more time

Answer 5

I'd recommend taking the minus out to avoid confusion

Answer 6

Okay, thanks. Where does the 2 (at v) come from?

Answer 7

You integrate e^(-x/2) you'll get\[\huge \int\limits e^{-x/2}dx \\ \\ \huge =\frac{e^{-x/2}}{(-\frac{1}{2})} \\ \\ \huge =-2e^{-x/2}\]

Answer 8

Thanks a lot!