Proving that the Yamabe invariant is indeed conformally invariant In Aubin's book, $\textit{Some Nonlinear Problems in Riemannian Geometry}$, the following proof is given to show that $\mu$ (the Yamabe invariant) is conformally invariant.

The Equation (1) that is being referred to is:

I have encountered two difficulties with this proof.
First: When I evaluate $\nabla^i(\varphi\psi)\nabla_i(\varphi\psi)$, I get
\begin{equation*}
\nabla^i(\varphi\psi)\nabla_i(\varphi\psi) = \varphi^2 \nabla^i\psi\nabla_i\psi + \psi^2\nabla^i\varphi\nabla_i\varphi + \varphi\psi(\nabla^i\varphi\nabla_i\psi + \nabla^i\psi\nabla_i\varphi).
\end{equation*}
How do the last three terms in the above turn into $\varphi\psi^2\Delta\varphi$?
Second: After using equation (1), I get,
\begin{equation*}
J(\varphi\psi)=\frac{4\frac{n-1}{n-2}\big[\int_M \varphi^2 \nabla^i\psi\nabla_i\psi \, dV\big] + \int_M R' \psi^2 dV'}{\big[\int_M \psi^N dV'\big]^{2/N}}.
\end{equation*}
In the first integral on the numerator, I'm not sure how $\varphi^2$ is used to transform $dV$ into $dV'$.
Apologies if the answers to my difficulties are trivial (I'm still rather new to this type of material). Any help would be greatly appreciated. Thank you.
 A: You're going to have to integrate by parts. Note also that Aubin uses the incorrect sign for $\Delta$, i.e. $\Delta u=-\mathrm{div}\,\nabla u$.
Integrating the second term by parts gives
$$\int\psi^2|\nabla\varphi|^2=-\int\varphi\,\mathrm{div}(\psi^2\nabla\varphi)=\int\varphi\psi^2\Delta\psi-2\int\varphi\psi\nabla\psi\cdot\nabla\varphi.\tag{$*$}$$
Now notice that the last two terms in what you wrote are actually the same and equal together $2\varphi\psi\nabla\psi\cdot\nabla\varphi$. This cancels with the second term on the right in $(*)$.
As for the volume element, in local coordinates $dV=\sqrt{\det g}\,dx.$ Now we have $$\det g'=\det(\varphi^{4/(n-2)}g)=\varphi^{4n/(n-2)}\det g.$$
It follows that $dV'=\varphi^{2n/(n-2)}dV$. This looks wrong, but that's because you also have to account for the difference between $\nabla \psi\cdot\nabla\varphi$ and $\nabla'\psi\cdot\nabla'\varphi$. There's another factor of $\varphi$ that comes from this because of the conformal change in metric. If you work this out, you get 
$$\int \nabla'u\cdot\nabla'v\,dV'=\int\nabla u\cdot\nabla v\,\varphi^2\,dV$$
for any smooth functions $u$ and $v$. This should get you what you need. 
