How Do You Solve Integrals of the Form exp(ax^q) When q Is Not 1?

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Please help. How do you solve this
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This seems to involve the elliptical integral according to maple...
 
Hi,
First, separate out the easy parts; write the integrand as:

(1/e2)e1/(2x) + 4x

Integrate 4x separately, = 2x2

Lookup the integral of e1/(2x) (! I found it on Mathematica, here:

http://integrals.wolfram.com/index.jsp?expr=E^(1/(2x))&random=false"

!)

Your result will be (1/e2)[The integral you found...] + 2x2

BTW: Wolfram/Mathematica does find that the answer includes a so-called "Exponential Integral".
 
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Oops I meant exponential not elliptical :)
 
The bigger question is how to solve an integral exp(ax^q) where q isn't null and isn't 1, I don't think there's a prescription to it.
are you sure it shouldn't be xexp(1/x), cause this can be computed by changing 1/x=u and a series change.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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