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Mate, it was 3am when I posted that, I don't have a clue what I was writing 2e^(x-3) should be the correct answer, according to integrals.wolfram.com.

Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

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Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

None of them.

(e^(x-3))*2

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Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

None of them.

(e^(x-3))*2

That's easy, it stays exactly the same (plus a certain constant C).

Yep, I definitely understood all of that.

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Depends, if you meant (e^(x-3))^2 which is equal to e^(2(x-3)) then the antiderivative would be 2e^(2(x-3)). But if you meant e^((x-3)^2) then you'll need to use the power series expansion similar to this: http://www.physicsforums.com/showthread.php?t=127015

Wtf !!!

post-6387-0-92889000-1369328469_thumb.jp

:Goober:

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Well, they have what entrepreneurs want + they deal with the money. Everyone wants what they have + we pay for a good one.

Yes, but do you agree that if entrepreneurs (or whoever) is paying them 'that much' money, it is because they make a lot more? :eyebrows:

So...no golden pussy for accountants!

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Yes, but do you agree that if entrepreneurs (or whoever) is paying them 'that much' money, it is because they make a lot more? :eyebrows:

So...no golden pussy for accountants!

Yessssss

They have 'undergound pussy' , man!!

Yesssss probably the wildest and the best .....

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