How do I find the indefinite integral to $\displaystyle \int e^{\sqrt x}dx$?

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- Sep 12th 2009, 02:09 PMchengbinIntegral
How do I find the indefinite integral to $\displaystyle \int e^{\sqrt x}dx$?

- Sep 12th 2009, 02:51 PMpickslides
Consider the following

$\displaystyle

\int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c

$ - Sep 12th 2009, 03:04 PMchengbin
I must be an idiot, because I dismissed that idea immediately without even trying!

Maybe it was the star beside the question that screwed up my mind. - Sep 12th 2009, 03:07 PMTWiX
u = √x ---> uČ = x ---> 2 u du = dx

substitute then use integration by parts .. - Sep 12th 2009, 03:08 PMpickslides
Radicals do have a strange way of throwing people off track

- Sep 12th 2009, 03:13 PMPlato
Differentiate $\displaystyle 2e^{\sqrt x } \sqrt x - 2e^{\sqrt x } $.

What do you get? - Sep 12th 2009, 03:19 PMskeeter
- Sep 12th 2009, 03:35 PMTWiX
- Sep 12th 2009, 04:09 PMJG89
- Sep 12th 2009, 05:15 PMKrizalid
- Sep 13th 2009, 01:57 AMpickslides