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Simple Antiderivative 
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#1
Oct505, 10:31 PM

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How would I compute the antiderivative of
[tex]\int \sqrt{1\frac{x^2}{2}}[/tex] It looks familiar, but I can't quite remember how... 


#2
Oct505, 10:46 PM

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PF Gold
P: 562

Put
[tex]x=\sqrt{2} \sin u[/tex] and go from there. 


#3
Oct605, 03:14 AM

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U can also put the Riemann measure on [itex]\mathbb{R} [/itex] : [itex] dx [/itex].
Daniel. 


#4
Oct605, 01:16 PM

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Simple Antiderivative
Ah yes, of course. Thanks.
Quick followup: [tex]\int\log\sqrt{1x^2}+x[/tex] 


#5
Oct605, 09:59 PM

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I think that integrand is related to an inverse hyperbolic trig function... but I'd have to play around with it to work out which one. Maybe somebody else...



#6
Oct605, 10:03 PM

P: n/a

Perhaps it simply cannot be expressed algebraically?



#7
Oct605, 10:07 PM

P: 666

I love Mathematica Alex 


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