这个积分怎么求出原函数?

令v=1-u,则S=∫(u+v+u^2*v^2)^(1/2)du=∫(u+v+u^2*v^2)^(1/2)dv“=”两边相乘,则S^2=(∫∫(u+v+u^2*v^2)dudv)=∫dv∫(u+v+u^2*v^2)du=∫(u^2/2+uv+u^3/3*v^2+C1)dv=(u^2/2*v+uv^2/2+u^3*v^3/9+C1v+C2)S=±√(u^2/2*v+uv^2/2+u^3*v^3/9+C1v+C2)=±√[u^2/2*(1-u)+u(1-u)^2/2+u^3*(1-u)^3/9+C1*(1-u)+C2]PS：不得不佩服楼上的证明,神来之笔!