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


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. 
