(1+y²)dx-x²(1+x²)ydy=0(1+y²)dx-1/2*x²(1+x²)d(y²)=0令u=y²,那么(1+u)dx=1/2*x²(1+x²)du∴2dx/[x²(1+x²)]=du/(1+u) 2[1/x²dx-1/(1+x²)dx]=d(1+u)/(1+u)两边同时积分,得:2(-1/x-arctanx)=ln(1+u)+C即ln(1+u)+2arctanx+(2/x)+C=0也即:ln(1+y²)+2arctanx+(2/x)+C=0望采纳