Just to close this off: note that $d=\partial+\bar\partial$ and that $\partial^2=0$ so $\partial\bar\partial=d\bar\partial$, and therefore $\Omega+\partial\bar\partial\varphi=\Omega+d\bar\partial\varphi$ is in the same cohomology class as $\Omega$. Since wedge product of forms descends to the usual product in cohomology, $(\Omega+d\bar\partial\varphi)^n=\Omega^n$ in cohomology, giving the same volume integral over our compact manifold.
Ben McKay
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