For $2 \le T \le x$, can we bound $$\int_1^{1+\frac{1}{log x}}\frac{x^s}{s}\zeta(\sigma + iT) \, \mathrm{d}\sigma \ll \frac{x \log^3 x}{T}$$
Question Background
We get the expression for $Z(s)$ by writing the Dirichlet series corresponding to $\mu \star (\frac{\mu^2 \phi}{\text{id}})$ as an Euler product.
We then note that $Z(s)$ converges absolutely for $\mathfrak{R}(s) > \frac12$
Main Part of Question
We aim for the above expression by considering the contour integral over the rectangle with corners at $(\frac12 + \epsilon \pm iT)$, $(c \pm iT)$.
The first term on the RHS is just the residue of the $\frac{x^s}{s} \zeta(s) Z(s)$ at its pole at $s = 1$.
The first error term on the RHS bounds the integral along the left vertical path (from $(\frac12 + \epsilon + iT)$ to $(\frac12 + \epsilon - iT)$).
The LHS and the second error term on the RHS just comes from part (i), which is the integral over the right vertical path.
However, I am a bit stuck on how to bound the horizontal integrals. Specifically, for $\sigma \le 1$, we can bound $\zeta(\sigma + iT) \ll |T|^{1-\sigma} \log|T|$; this leaves the part of the horizontal integral from $\sigma = 1$ to $\sigma = c = 1 + \frac{1}{log x}$, on which I am stuck as I am unsure of how to bound $\zeta(\sigma + i T)$. (The other terms are easy to bound on this part of the path - $Z(s) \le Z(\frac12 + \epsilon) = \mathcal{O}(1)$, for example.)
We need to bound the horizontal integrals by either of the two error terms; it seems, that we stand more chance of bounding it by the second error term, $\mathcal{O}(\frac{x \log^3 x}{T}), and in order to do this, it seems that we need
$$\int_1^{1+\frac{1}{log x}}\frac{x^s}{s}\zeta(\sigma + iT) \, \mathrm{d}\sigma \ll \frac{x \log^3 x}{T}$$
Edit: Based on the comment by Daniel Fischer, it seems that we have $\zeta(\sigma + iT) = \mathcal{O}(\log T)$ on $\sigma > 1$ (and obviously this suffices for the purposes of the original question).
Edit 2: As explained by Daniel Fischer in the comments, this bound is proved here.