Proof in Hamilton: Divergence theorem for differential forms? For a vector field $X\in\Gamma(TM)$ on a closed Riemannian manifold $(M,g)$ we have
\begin{align*}
\int_M\text{div}X\;\mu=0,
\end{align*}
where
\begin{align*}
\text{div}X=-g^{ij}g(\nabla_iX,\partial_j).
\end{align*}
Here $\mu$ is the volume form and $\nabla$ is the Levi-Civita connection of the metric $g$.
Question:  Is there an analogous statement for differential forms and even for general tensors?
In other words if $\omega\in\Gamma(T^*M)$ is a $1$-form does it hold that
\begin{align*}
\int_M\text{div}\,\omega\;\mu=0\,?
\end{align*}
Here
\begin{align*}
\text{div}\,\omega=-g^{ij}\nabla_i\omega_j.
\end{align*}
We can easily define the divergence for higher-order tensors, so does the 'divergence theorem' also hold here as well?
EDIT:  The motivation for this question is as follows.  Let $M$ be as above and
\begin{align*}
\dot{R}_{jk}=D\text{Rc}(g)h_{jk}=\frac{1}{2}g^{pq}(\nabla_q\nabla_jh_{pk}+\nabla_q\nabla_kh_{pj}-\nabla_q\nabla_ph_{jk}-\nabla_j\nabla_kh_{pq})
\end{align*}
be the linearisation of the Ricci tensor at $g$ in the direction of the symmetric tensor $h$.  Then there is a statement in a paper by Hamilton that 
\begin{align*}
\int_M\text{tr}_g\dot R_{jk}\,\mu=\int_Mg^{jk}\dot R_{jk}\,\mu=0.
\end{align*}
I calculate that
\begin{align*}
\text{tr}_g\dot R_{jk}=g^{jk}\dot R_{jk}&=g^{jk}\frac{1}{2}g^{pq}(\nabla_q\nabla_jh_{pk}+\nabla_q\nabla_kh_{pj}-\nabla_q\nabla_ph_{jk}-\nabla_j\nabla_kh_{pq})\\
&=g^{jk}g^{pq}\nabla_q\nabla_jh_{pk}-g^{jk}g^{pq}\nabla_q\nabla_ph_{jk}\\
&=\text{div}(\text{div}\,h)-\Delta\text{tr}_gh.
\end{align*}
The divergence $(\text{div}\,h)_p=g^{jk}\nabla_jh_{pk}$ of the symmetric tensor $h$ is a $1$-form.  
QUESTION:  The integral of the connection Laplacian is zero but how is the integral of the other term zero?  Or am I doing something silly and wrong...?
 A: On a Riemannian manifold, the divergence theorem applies to $1$-forms as well as to vector fields. The simplest way to see this is by using the "musical isomorphisms" between $1$-forms and vector fields. This are the inverse isomorphisms $\flat\colon TM\to T^*M$ and $\sharp\colon T^*M\to TM$ defined by raising and lowering indices. If $X$ is a vector field and $\omega$ is a $1$-form, 
$$
(X^\flat)_i = g_{ij}X^j, \qquad 
(\omega^\sharp)^j = g^{jk}\omega_i.
$$
Then the divergence of $\omega$ is simply defined to be $\operatorname{div}\omega := \operatorname{div}(\omega^\sharp)$. In components, $\operatorname{div}\omega = g^{jk}\nabla_j\omega_k$.
The divergence theorem for $1$-forms then follows directly from the one for vector fields:
$$
\int_M \operatorname{div}\omega\, \mu = 
\int_M \operatorname{div}(\omega^\sharp)\, \mu = 
0.
$$
A: There is no Stokes's Theorem for general tensors. The general result is that $\displaystyle\int_M d\omega = 0$ for any compact oriented $n$-dimensional manifold $M$ without boundary and $(n-1)$-form $\omega$ on $M$. The divergence theorem you cite is for the case of $\omega = \star\eta$ where $\eta$ is the $1$-form corresponding to $X$ under the isomorphism $T^*M\cong TM$ induced by the Riemannian metric.
A: Addition to Jack Lee's answer:  Since $\nabla_i\omega^\sharp=(\nabla_i\omega)^\sharp$ (the covariant derivative respects the musical isomorphisms) and we also calculate that
\begin{align*}
\text{div}\,\omega&:=\text{div}\,\omega^\sharp\\
&=-g^{ij}g(\nabla_i\omega^\sharp,\partial_j)\\
&=-g^{ij}((\nabla_i\omega)^\sharp)^\flat_j\\
&=-g^{ij}\nabla_i\omega_j.
\end{align*}
