Question

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

Answer 1

Have a look:

Answer 2

`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`