integration by parts problem

[afcwestwarrior]

1. The problem statement, all variables and given/known data
∫e^x+e^x




2. Relevant equations
∫u dv= uv- ∫v du


3. The attempt at a solution
u= x+e^x
du= e^x

so it would be e^u

integral = e^u

= e^(e^x) +c is that correct, i know the answer is but what i just did

[Defennder]

Did you mistype the integral? \int 2e^x dx?

[HallsofIvy]

1. The problem statement, all variables and given/known data
∫e^x+e^x[quote]
??? Surely you don't mean 2\int e^x dx?




[quote]2. Relevant equations
∫u dv= uv- ∫v du


3. The attempt at a solution
u= x+e^x
du= e^x
Are you now saying the problem is \int (xe^x+ e^x)dx?
Then you are not using integration by parts, you are using a simple substitution. Yes, \int (xe^x+ e^x)dx= \int (x+ e^x)e^x dx. If you let u= x+ ex, then du= (1+ e^x) dx, not just ex dx. And please by sure to include the "dx" in the integral; that may be part of what is confusing you.

so it would be e^u

integral = e^u
= e^(e^x) +c is that correct, i know the answer is but what i just did
Well, you can always check an integration yourself by differentiating.
\frac{d}{dx}\left(e^{e^x}\right)= \frac{de^u}{du}\frac{de^x}{dx}
with u= ex
= (e^u)(e^x)= (e^{e^x})(e^x)
Which is not what you started with.
Did you consider just doing the two integrals separately?
[tex]\int (xe^x+ e^x)dx= \int xe^x dx+ \int e^x dx[/itex]
You should be able to do the second of those directly and the first is a simple integration by parts.