How to derive this equality when deducing Kelvin's Theorem?

Question

I can't understand how to obtain this equality from the deduction of Kelvin's Theorem: $$\oint_C\boldsymbol{u}\cdot(\mathrm{d}\boldsymbol{s}\cdot\nabla)\boldsymbol{u}=\frac{1}{2}\oint_C\nabla(|\boldsymbol{u}|^2)\cdot\mathrm{d}\boldsymbol{s}$$ where $\boldsymbol{u}$ is the velocity vector and $\mathrm{d}\boldsymbol{s}$ is an element along the closed contour. I have known about this vector identity: $$\boldsymbol{u}\cdot\nabla\boldsymbol{u}=\nabla(|\boldsymbol{u}|^2/2)-\boldsymbol{u}\times(\nabla\times\boldsymbol{u})$$ Combine these two equalities and I will obtain this, $$\oint_C\boldsymbol{u}\times(\nabla\times\boldsymbol{u})\cdot\mathrm{d}\boldsymbol{s}=0$$ Why on earth could this term be zero???