Integrate dx/(1-x): Solve (2/3a)(1-(1-aW)^(3/2))

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Thread starter cycling4life
Start date Jan 31, 2012
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cycling4life

Thread starter
Jan 31, 2012

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-"?

Last edited: Jan 31, 2012

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JJacquelin

Feb 1, 2012

-ln(1-x) = (-2/3a)*(1-aW)^(3/2) + Contant

Integrate dx/(1-x): Solve (2/3a)(1-(1-aW)^(3/2))

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Integrate dx/(1-x): Solve (2/3a)(1-(1-aW)^(3/2))

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