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

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SUMMARY

The discussion centers on the integration of the function dx/(1-x) and its relationship to the expression (1-aW)^(1/2) dW. The user derives -ln(1-x) = (-2/3a)*(1-aW)^(3/2) but questions the additional "1-" in the alternative expression ln(1/(1-x)) = (2/3a)((1-(1-aW)^(3/2)). The key conclusion is that the presence of the "1-" is due to the logarithmic identity that transforms the expression into a different form, maintaining equivalence through the integration process.

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cycling4life
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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-"?
 
Last edited:
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-ln(1-x) = (-2/3a)*(1-aW)^(3/2) + Contant
 

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