How to show that $\int_M(\Delta f-|\nabla f|^2)(2\Delta f-|\nabla f|^2)e^{-f} dV=\int_M-\nabla_if\nabla_i(2\Delta f-|\nabla f|^2)e^{-f}dV$

Question

$M$ is a compact Riemannian manifold. $f$ is function on $M$

How to show that $\int_M(\Delta f-|\nabla f|^2)(2\Delta f-|\nabla f|^2)e^{-f} dV=\int_M-\nabla_if\nabla_i(2\Delta f-|\nabla f|^2)e^{-f}dV$?

I feel I should use integration by parts, but failed.

Thanks for any useful hint or answer.

Below picture is from 201th page of this paper.

Answer 1

This is integration by part, (or divergence theorem):

Write $H = 2\Delta f-|\nabla f|^2$, then $$\begin{split} \int_M \Delta f H e^{-f} dV & = -\int_M \nabla f\cdot \nabla(He^{-f} )dV \\ &= -\int_M \nabla f\cdot (e^{-f}\nabla H+ H \nabla (e^{-f}) )dV \\ &= -\int_M \nabla f \cdot \nabla H e^{-f} dV - \int_M \nabla f \cdot (e^{-f}( -\nabla f))H dV \\ &= -\int_M \nabla f \cdot \nabla H e^{-f} dV + \int_M |\nabla f|^2 H e^{-f} dV \end{split}$$

Move the second term to the left and you are done.