# Integration by parts with a simple substitution

• Sep 3rd 2010, 06:26 PM
houndmom87
Integration by parts with a simple substitution
Integrate: (1/3)(x)[e^(x^1/3)]dx

I have tried making u=1/3x and dv=e^(x^1/3)
I've also tried to do a simple substitution first. I keep getting stuck and I've tried every combination I can think of. I would greatly appreciate any help. Thanks so much!
• Sep 3rd 2010, 07:25 PM
mr fantastic
Quote:

Originally Posted by houndmom87
Integrate: (1/3)(x)[e^(x^1/3)]dx

I have tried making u=1/3x and dv=e^(x^1/3)
I've also tried to do a simple substitution first. I keep getting stuck and I've tried every combination I can think of. I would greatly appreciate any help. Thanks so much!

integrate &#40;x&#47;3&#41;Exp&#91;x&#94;&#40;1&#47;3&#41;&# 93; - Wolfram|Alpha (Click on show steps)
• Sep 3rd 2010, 07:59 PM
houndmom87
I guess I'm just unsure as to how to get u^5. Thanks!
• Sep 4th 2010, 02:07 PM
mr fantastic
Quote:

Originally Posted by houndmom87
I guess I'm just unsure as to how to get u^5. Thanks!

Have you tried doing the algebra? Please show your attempt with this calculation.
• Sep 5th 2010, 05:27 PM
HallsofIvy
Quote:

Originally Posted by houndmom87
Integrate: (1/3)(x)[e^(x^1/3)]dx

I have tried making u=1/3x and dv=e^(x^1/3)
I've also tried to do a simple substitution first. I keep getting stuck and I've tried every combination I can think of. I would greatly appreciate any help. Thanks so much!

But it isn't the "1/3x" that is the problem! It is that "x^1/3" in the exponential. I recommend letting u= x^{1/3}. Then du= (1/3)x^{-2/3}dx so that dx= 3x^{2/3}du. That way (1/3)xdx= (1/3)x(3x^{2/3}du)= x^{5/3}du. And since u= x^(1/3), x^(5/3)= (x^(1/3))^5= u^5.

That is, (1/3)xdx= u^5du and so (1/3)xe^{x^1/3}dx= u^5e^udu.

Now that's still going to be a nuisance- use "integration by parts"- five times!