1/2*∫d(y²+1)/(y²+1)=1/2*∫[1/(x-1)-1/(x+1)]dx
ln(y²+1)=ln|x-1|-ln|x+1|+lnC=lnC|(x-1)/(x+1)|
y²+1=C|(x-1)/(x+1)|
y=√[C|(x-1)/(x+1)|-1]