- #1

- 22

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter maobadi
- Start date

- #1

- 22

- 0

- #2

- 623

- 0

This seems to involve the elliptical integral according to maple...

- #3

- 12

- 0

Hi,

First, separate out the easy parts; write the integrand as:

(1/e^{2})e^{1/(2x)} + 4x

Integrate 4x separately, = 2x^{2}

Lookup the integral of e^{1/(2x)} (! I found it on Mathematica, here:

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

!)

Your result will be (1/e^{2})[The integral you found...] + 2x^{2}

BTW: Wolfram/Mathematica does find that the answer includes a so-called "Exponential Integral".

First, separate out the easy parts; write the integrand as:

(1/e

Integrate 4x separately, = 2x

Lookup the integral of e

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

!)

Your result will be (1/e

BTW: Wolfram/Mathematica does find that the answer includes a so-called "Exponential Integral".

Last edited by a moderator:

- #4

- 623

- 0

Oops I meant exponential not elliptical :)

- #5

MathematicalPhysicist

Gold Member

- 4,460

- 274

are you sure it shouldn't be xexp(1/x), cause this can be computed by changing 1/x=u and a series change.

Share: