Prove that, ${\LARGE\int}_M d^4x\sqrt{|g|}A_{\mu\nu}(D^\mu B^\nu) = {\LARGE\oint}_{\partial M} d^3\sigma^\mu A_{\mu \nu}B^\nu - {\LARGE\int}_M d^4x\sqrt{|g|}(D^\mu A_{\mu \nu})B^\nu$, where $\partial M$ is the three-dimensional boundary of a four-dimensional manifold $M$; $d^3\sigma^\mu$ is a four dimensional vector perpendicular to $\partial M$, directing inside $M$, whose modulus is equal to the volume element of $\partial M$.
I have absolutely no idea how to tackle this one, all I have managed to figure out that $d^4x$ here signifies the volume element of four-dimensional space $dx^1dx^2dx^3dx^4$, and $D^\mu$ is the standard covariant differential operator, also $|g|$ is the determinant of metric tensor. But how can I use these ideas to prove the above integral ? Any help will be appreciated in this regard, and any physical interpretation of that problem will be also very helpful to me.
You can use the generalized Stoke's theorem to show this. First, note that
$$ \color{blue}{D^\mu(A_{\mu\nu}B^{\nu})} = \color{red}{(D^\mu A_{\mu\nu} )B^\nu} + \color{orange}{A_{\mu\nu}(D^\mu B^\nu)} $$
So that
\begin{eqnarray} \int_M {\rm d}^4 x~ \sqrt{|g|}\color{orange}{A_{\mu\nu}(D^\mu B^\nu)} &=& \int_M {\rm d}^4 x~ \sqrt{|g|}\color{blue}{D^\mu(A_{\mu\nu}B^{\nu})} -\int_M {\rm d}^4 x~ \sqrt{|g|}\color{red}{(D^\mu A_{\mu\nu} )B^\nu} \\ &=& \int_{\partial M} {\rm d}^3 x \sqrt{|h|} \sigma^\mu (A_{\mu\nu}B^{\nu}) - \int_M {\rm d}^4 x~ \sqrt{|g|}(D^\mu A_{\mu\nu} )B^\nu \end{eqnarray}
where $h$ represents the determinant of the metric on $\partial M$ resulting from pulling back the metric on $M$. For more details go to Eq. (9.68) of this reference