A two-dimensional Pfaff form with a global integrating factor

The first derivatives of the coefficients $2y$ and $1+x$ are $0,2$ and $1,0$, respectively, and all their second partial derivatives vanish; hence $\omega$ is $C^2$ and nowhere zero. Expanding $S(x,y)=(1+x{)}^{2}y$ gives $S(x+h,y+k)-S(x,y)=2y(1+x)h+(1+x{)}^{2}k+2(1+x)hk+y{h}^2+h^{2}k$. The last three terms divided by $\sqrt{h^2+k^2}$ tend to zero as $(h,k)\to (0,0)$, so $d{S}_{(x,y)}(h,k)=2y(1+x)h+(1+x{)}^{2}k$. These first derivative coefficients have continuous first partials $S_{xx}=2y$, $S_{xy}=S_{yx}=2(1+x)$ and $S_{yy}=0$, so $S\in C^2$. Thus $dS=2y(1+x)dx+(1+x{)}^{2}dy=(1+x)\omega =\mu \omega$. Because $1+x>0$ on $U$, both $\mu$ and $dS$ are nowhere zero.

For any piecewise-$C^1$ horizontal path $\gamma$, the identity in step 1.1 and [F1]–[F2] give $(S\circ \gamma {)}^{\mathrm{\prime }}=dS(\dot{\gamma })=\mu \omega (\dot{\gamma })=0$ on the interior of each smooth piece. Thus every horizontal tangent lies in $\ker dS=\ker \omega$ along that piece.