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

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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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-ln(1-x) = (-2/3a)*(1-aW)^(3/2) + Contant
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

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