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-"?
Re: Integration question
-ln(1-x) = (-2/3a)*(1-aW)^(3/2) + Contant
|All times are GMT -5. The time now is 01:45 AM.|
Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums