I think there is a summation in the integral. For example, $\partial_i^2 u$ should be $\sum_{i=1}^n\partial_i^2 u=\Delta u$, and $\partial_k^2 u$ should be $\sum_{k=1}^n\partial_k^2 u=\Delta u$, etc. With this understanding, the formula you are asking should be $\int_\Omega(\Delta u)^2=\int_{\partial\Omega}(\Delta u)\nabla u\cdot n dS-\int_\Omega\nabla u\cdot\nabla(\Delta u).$ To derive this, one can apply the divergence theorem to the vector field $X=(\Delta u)\nabla u$. That is: $\int_\Omega\nabla\cdot X=\int_{\partial\Omega}X\cdot ndS.$ Then $LHS=\int_\Omega\nabla\cdot X=\int_\Omega\nabla\cdot ((\Delta u)\nabla u)=\int_\Omega\nabla u\cdot\nabla(\Delta u)+\int_\Omega(\Delta u)^2$ because $\nabla\cdot\nabla u=\Delta u$. On the other hand, $RHS=\int_{\partial\Omega}X\cdot ndS=\int_{\partial\Omega}(\Delta u)\nabla u\cdot ndS.$