How do I find the indefinite integral to $\displaystyle \int e^{\sqrt x}dx$?
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Consider the following $\displaystyle \int e^{f( x)}dx = \frac{e^{f( x)}}{f'( x)}+c $
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.
u = √x ---> uČ = x ---> 2 u du = dx substitute then use integration by parts ..
Radicals do have a strange way of throwing people off track
Differentiate $\displaystyle 2e^{\sqrt x } \sqrt x - 2e^{\sqrt x } $. What do you get?
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 ... $\displaystyle 2e^u(u - 1) + C$ $\displaystyle 2e^{\sqrt{t}}(\sqrt{t} - 1) + C$
Originally Posted by pickslides Radicals do have a strange way of throwing people off track Is this for me ?
Originally Posted by TWiX Is this for me ? I'd say no, seeing as your method was correct.
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.
Originally Posted by Krizalid that's a wrong formula. 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.
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