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How do you integrate #1/(y(1-y))#?

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May 18, 2015

Have a look:
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May 18, 2015

#int1/(y(1-y))dy=-int1/(y^2-y+(-1/2)^2-(-1/2)^2)dy#

#=>-int1/((y-1/2)^2-(1/2)^2)dy=-1/(1/2)^2int1/(((y-1/2)/(1/2))^2-1)dy=-4int1/((2y-1)^2-1)dy#

#=>1/2*-4artanh(2y-1) + C = -ln|(1-2y+1)/(1+2y-1)| + C=ln|(y)/(1-y)| + C#

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