I've been looking at this breakdown that deals with taking the gradient of an expectation w.r.t the distribution the expectation is sampling from. On equation 2 is where I get stuck. Where does the $+1$ on the RHS of eq (2) go? I was/am expecting it to be: $\int{ (\log p_\theta(y) - \log q_\phi(x) +1 ) \nabla_{\phi} q_\phi(x) dz }$.
I've my answer and should add that both $p(x,z)$ and $q_\phi (z|x)$ are probabilities. By integrating over one (to get 1), then taking the gradient (to get 0), my $+1$ dissapears.
$$\begin{align} \nabla_{\phi} E_{q_\phi(z|x)} \left [ \log p(x, z) - \log q_\phi(z|x) \right] &= \nabla_{\phi} \int{\log p(x, z) q_\phi(z|x) dz} -\nabla_{\phi} \int{q_\phi(z|x) \log q_\phi(z|x) dz } \tag{1}\\ &=\int{\log p(x, z) \nabla_{\phi} q_\phi(z|x) dz} - \int{(\log q_\phi(z|x) +1) \nabla_{\phi} q_\phi(z|x) dz } \tag{2}\\ &= \int{ (\log p(x, z) - \log q_\phi(z|x) ) \nabla_{\phi} q_\phi(z|x) dz } \tag{3} \end{align}$$