Integration by parts..

[Fizex]

We know the formula is \inline{\int udv=uv-\int vdu} but when you say that for example, dv=e^x dx, then why when you integrate to get v, you don't include the integration constant?

For this integral:
\int xe^{x}dx
dv = e^x dx
v = e^x + C?

[CompuChip]

You can, in this case you would get
\int x e^x \, \mathrm dx = (e^x + C) x + \int (e^x + C) \, \mathrm dx = x (e^x + C) - (e^x + C x + C')
If you expand
x e^x + C x - (e^x - C x + C') = (x - 1) e^x - C'

[Fizex]

oh, haha, I was only paying attention to one side of the equation. Thanks.