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Integration question

by cycling4life
Tags: integration
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cycling4life
#1
Jan31-12, 09:49 PM
P: 7
In part of a derivation they have, Integrate dx/(1-x) = (1-aW)^(1/2) dW

I get -ln(1-x) = (-2/3a)*(1-aW)^(3/2) but they say it's ln(1/(1-X)) = (2/3a)((1-(1-AW)^(3/2))

Can anyone tell me how they get that extra "1-"?
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JJacquelin
#2
Feb1-12, 02:43 AM
P: 756
-ln(1-x) = (-2/3a)*(1-aW)^(3/2) + Contant


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