# Integral

• Sep 12th 2009, 02:09 PM
chengbin
Integral
How do I find the indefinite integral to $\displaystyle \int e^{\sqrt x}dx$?
• Sep 12th 2009, 02:51 PM
pickslides
Consider the following

$\displaystyle \int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c$
• Sep 12th 2009, 03:04 PM
chengbin
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 PM
TWiX
u = √x ---> uČ = x ---> 2 u du = dx
substitute then use integration by parts ..
• Sep 12th 2009, 03:08 PM
pickslides
Radicals do have a strange way of throwing people off track
• Sep 12th 2009, 03:13 PM
Plato
Differentiate $\displaystyle 2e^{\sqrt x } \sqrt x - 2e^{\sqrt x }$.
What do you get?
• Sep 12th 2009, 03:19 PM
skeeter
Quote:

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

$\displaystyle u = \sqrt{t}$

$\displaystyle du = \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2u} \, dt$

$\displaystyle dt = 2u \, du$

$\displaystyle 2\int ue^u \, du$

integrate by parts ...

$\displaystyle 2e^u(u - 1) + C$

$\displaystyle 2e^{\sqrt{t}}(\sqrt{t} - 1) + C$
• Sep 12th 2009, 03:35 PM
TWiX
Quote:

Originally Posted by pickslides
Radicals do have a strange way of throwing people off track

Is this for me ?
• Sep 12th 2009, 04:09 PM
JG89
Quote:

Originally Posted by TWiX
Is this for me ?

I'd say no, seeing as your method was correct. (Thinking)
• Sep 12th 2009, 05:15 PM
Krizalid
Quote:

Originally Posted by pickslides
Consider the following

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

that's a wrong formula. (Sleepy)
• Sep 13th 2009, 01:57 AM
pickslides
Quote:

Originally Posted by Krizalid
that's a wrong formula. (Sleepy)

yep it is, was confused with $\displaystyle y= e^{f(x)} \Rightarrow \frac{dy}{dx}= f'(x)e^{f(x)}$

Gentle reminder to myself not to post on MHF the morning after a big night on the sauce.