It's been a while since I did integrals, so just figure out what you'd have to derive to get e^x².

The way I see it, you'd have to derive (e^x²)/(2x) to get e^x². Just use the chain rule to verify that.

SO, the integral would be (e^x²)/(2x) + K, where K is some constant (remember, constants get dropped when you do derivatives).

Hope that helps.

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Koska Kintaro stumbled drunkenly to the keyboard and typed:
See, that's what I figured, but my calculus prof. says the antiderivative of e^x² is insolvable in terms of variables.

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I think your prof is fucking with you. Smack him upside the head for me, OK?

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And I was all like 'Oh yeah?' and Karnaj was all like:
I think your prof is fucking with you. Smack him upside the head for me, OK?

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I dunno.

The derivative of (e^x²)/(2x) works out to be plain ol (e^x)/x, does it not?

[ 11-08-2001: Message edited by: Koska Kintaro ]

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Koska Kintaro had this to say about Matthew Broderick:
I dunno.

The derivative of (e^x²)/(2x) works out to be plain ol (e^x)/x, does it not?

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Oh shoot, that's right. You're prof's right, it's unsolvable.

But he's still fookin' wit j00.

That'd be why I'm in calc 1.

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Karnaj had this to say about Robocop:
You're prof's right, it's unsolvable.

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I propose a new constant: The Maradon (expressed as *M* with the stars like that)

It connotates the antiderivative of e^x²

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Koska Kintaro had this to say about Optimus Prime:
The derivative of (e^x²)/(2x) works out to be plain ol (e^x)/x, does it not?

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You forgot to carry the 1.

anti derivitive of plain e^x = e^x+c so that
if you let u=x^2 and
du/dx= x^3/3

it should come out to be

e^x^2=((x^3)/3)*e^x^2 +c

thought id try also this isnt a bad site either for some stuff
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/index.html

Goodluck

Morathu