1/(y^2-1) = (1/2)[1/(y-1) -1/(y+1)]∫dy/(y^2-1)=(1/2)∫[1/(y-1) -1/(y+1)] dy=(1/2)ln| (y-1)/(y+1) | + C
∫1/(y^2-1)dy=1/2∫[1/(y-1)-1/(y+1)]dy=1/2ln[(y-1)/(y+1)]+C
