Integral

**chengbin** · September 12th 2009, 02:09 PM

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

**pickslides** · September 12th 2009, 02:51 PM

Consider the following

$<br /> \int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c<br />$

**chengbin** · September 12th 2009, 03:04 PM

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.

**TWiX** · September 12th 2009, 03:07 PM

u = √x ---> u² = x ---> 2 u du = dx
substitute then use integration by parts ..

**pickslides** · September 12th 2009, 03:08 PM

Radicals do have a strange way of throwing people off track

**Plato** · September 12th 2009, 03:13 PM

Differentiate $2e^{\sqrt x } \sqrt x - 2e^{\sqrt x }$ .
What do you get?

**skeeter** · September 12th 2009, 03:19 PM

Originally Posted by chengbin

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

$u = \sqrt{t}$

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

$dt = 2u \, du$

$2\int ue^u \, du$

integrate by parts ...

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

$2e^{\sqrt{t}}(\sqrt{t} - 1) + C$

**TWiX** · September 12th 2009, 03:35 PM

Originally Posted by pickslides

Radicals do have a strange way of throwing people off track

Is this for me ?

**JG89** · September 12th 2009, 04:09 PM

Originally Posted by TWiX

Is this for me ?

I'd say no, seeing as your method was correct.

**Krizalid** · September 12th 2009, 05:15 PM

Originally Posted by pickslides

Consider the following

$<br /> \int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c<br />$

that's a wrong formula.

**pickslides** · September 13th 2009, 01:57 AM

Originally Posted by Krizalid

that's a wrong formula.

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

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

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