Why is $\int_M i \partial\bar \partial u \wedge \Phi^{n-1}$ zero?

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I was studying vanishing theorems on holomorphic sections of holomorphic Hermitian vector bundles on Kähler manifolds. Here $(E,h)$ is a Hermitian holomorphic bundle on a compact Kähler manifold $M$. The book claims that the following integral is $0$.

$$ \int_{M}i\partial \bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1} = 0. $$ Here $\Phi$ is the Kähler form on the manifold $M$ and $h$ is the Hermitian metric on $E$ and $\xi$ is a global holomorphic section of $E$.

I'm not able to see why this integral should be $0$.

I'll be happy to provide more details.

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As suggested in the comment, one can write

\begin{align*} \partial \bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1} &= d \bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1} \\ &= d(\bar{\partial}h(\xi,\xi) \wedge \Phi^{n-1}) \end{align*}

Since $\partial^2 = 0$ and $\Phi$ is closed. Thus the integrand is exact and Stokes theorem implies that it's zero.