The discussion focuses on the correct approach to solving the integral ∫(xe^x + e^x)dx using integration techniques. The initial attempt incorrectly applied integration by parts, while the correct method involves recognizing that the integral can be split into two separate integrals: ∫xe^x dx and ∫e^x dx. The latter can be solved directly, while the former requires integration by parts. The final solution is derived by combining the results of these two integrals.
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afcwestwarrior said:Homework Statement
∫e^x+e^x? Surely you don't mean [itex]2\int e^x dx[/itex]?
Are you now saying the problem is [itex]\int (xe^x+ e^x)dx[/itex]?Homework Equations
∫u dv= uv- ∫v du
The Attempt at a Solution
u= x+e^x
du= e^x
Then you are not using integration by parts, you are using a simple substitution. Yes, [itex]\int (xe^x+ e^x)dx= \int (x+ e^x)e^x dx[/itex]. 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.
Well, you can always check an integration yourself by differentiating.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
[tex]\frac{d}{dx}\left(e^{e^x}\right)= \frac{de^u}{du}\frac{de^x}{dx}[/tex]
with u= ex
[tex]= (e^u)(e^x)= (e^{e^x})(e^x)[/tex]
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]<br /> You should be able to do the second of those directly and the first <b>is</b> a simple integration by parts.[/tex]