Integrate 1/(y-y^2) dy?

Answer 1

you must use partial fractions

A / y + B / (y-1) ===> A=-1 and B=1
Then it becomes
integral of -1/y + 1/(y-1)
can you do this?
- ln IyI + ln Iy-1I +C
which will simplify to ln I (y-1)/y I + C

:)

Answer 2

Int[1/(y-y^2) dy]=-Int[dy/(y^2-y)]=-Int[dy/((y-0.5)^2-1)]=
=-1/2[ln|(y-1.5)/(y+0.5)|+C.

Answer 3

Try partial fractions.