Integrating e^{\sqrt{x}}/\sqrt{x} from 1 to 4

[kartoshka]

$\int$ $\frac{e^{\sqrt{x}}}{\sqrt{x}}$


It's in the substitution rule/symmetric function section of my book, so I figure I probably have to use one of those techniques to solve it. I've tried doing a bunch of different u substitutions $\sqrt{x}$, $e^{{\sqrt{x}}}$, etc, but none of them seem right.

How can you tell if a function is symmetric by looking at the equation? And whether it is even or odd?

PS - couldn't figure out how to do it, but it's actually a definite integral that goes from 1 to 4. Also, if the top of the fraction is hard to read, it's $e^{{\sqrt{x}}}$.

 

[l'Hôpital]

Try $$u = \sqrt{x}$$ again.

 

[kartoshka]

Well now I feel a bit ridiculous. But I end up with $/int$e^u * u^-2. Is there a way to solve this without integration by parts? We haven't gotten to it yet so I feel like there should be a way.

Sorry for the lack of formatting, typing on my phone and I can't remember most of the tags.

 

[SammyS]

If $u = \sqrt{x}\,,$ then what is du ?

BTW: It's important to have the dx in the integral: $\displaystyle \int\frac{e^{\sqrt{x}}}{\sqrt{x}} dx\,.$