# ¿How do I integrate $e^\sqrt{x}$, using the substitution rule?

I've been trying to integrate that function, but it looks like I'm missing something, so can any one please show me how to integrate $e^\sqrt{x}$, in order to correct my procedure. Thanks

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If $x=u^2$, then $\mathrm dx=\dots$ – J. M. Sep 26 '11 at 21:53
Hint: substitute $u = \sqrt{x}$. – Robert Israel Sep 26 '11 at 21:54
No is not a homework I'd tried to do the same substitution that is shown down in the first answer, but then I don't know what to do then, since there is a product in the integrand. – Davynch0 Sep 26 '11 at 22:13

You're going to want to start with a $u$-substitution $u=\sqrt{x}$. Notice then that $u^2=x$, so $2udu=dx$. Then $$\int e^\sqrt{x}dx=2\int e^u u\ du.$$

You can then proceed by integration by parts.

To get started, you can set $w=u$ and $dv=e^u\ du$, and use the fact that $\int w\ dv= wv-\int v\ dw$. Lastly, don't forget to substitute back to get the integral in terms of $x$.

Continuing, $dw=du$ and $v=e^u$. Then $$\int w\ dv= wv-\int v\ dw = e^u u-\int e^u\ du=e^u u-e^u.$$ So $$\int e^\sqrt{x}dx=2\int e^u u\ du=2e^uu-2e^u+C=2e^\sqrt{x}(\sqrt{x}-1)+C$$ after back substituting $u=\sqrt{x}$ and adding a possible constant term.

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I don't know how to integrate by parts, can't it be solved by the substitution rule at all? – Davynch0 Sep 26 '11 at 22:07
@Dav Since you say it is not homework, I've tried to add a more complete solution. Hopefully it's clearer. – yunone Sep 26 '11 at 22:20
thanks a lot @yunone – Davynch0 Sep 26 '11 at 22:21
Not homework, and you don't know integration by parts? Maybe that is why it is not (yet) homework. Maybe it WILL be homework later, after you do learn integration by parts! – GEdgar Sep 27 '11 at 0:12
@Davyncho: It really is (after the natural substitution) a standard integration by parts. But if we insist on substitution alone, we can (I am joking, but it is true) use $u=(\sqrt{x}-1)e^\sqrt{x}$. – André Nicolas Sep 27 '11 at 1:37