求方程x(1+y^2)dx-y(1+x^2)dy=0的通解

x(1+y²)dx=y(1+x²)dy
x/(1+x²)dx=y/(1+y²)dy
2x/(1+x²)dx=2y/(1+y²)dy
1/(1+x²)d(1+x²)=1/(1+y²)d(1+y²)
ln(1+x²)=ln(1+y²)+lnc
1+x²=c(1+y²)