${\bf W.t.s:}$
I'd like to show for the generalised Laplacian defined on a compact, connected orientable n dimensional Riemannian manifold ($M, g$) ie. for $\Delta u = $div ($\nabla u$) that
$$\int_M u\Delta u = -\int_M {\lvert \nabla u \rvert}^2$$
${\bf Attempt:}$
I've tried using the identity
$$\mbox{div} (X) \Omega = \mathcal{L}_{X} \Omega$$
on $X = \nabla u$ and where $\Omega$ is the Riemannian volume form. This gives:
$$\int_M u\Delta u \Omega = \int_M u \mbox{div} (\nabla u)\Omega = \int_M u\mathcal{L}_{\nabla u} \Omega \underset{Cartan}{=} \int_M ud(i_{\nabla u} \Omega) + ui_{\nabla u}(d\Omega)$$.
It is here where I get stuck. There may well be a more efficient way of showing this so please feel free to suggest alternative methods.
In Euclidean space I know this would be a standard integration by parts. But I presume this is out of the question given that we are in a more general setting.