We know that $\frac{\zeta'}{\zeta}(s)=O(\log\vert s\vert) $ for $\mathrm{Re}(s)<-1$ and $\vert s+2m\vert >1/4$ for all $m\in\mathbb{N}$, what means that $s$ is not near to some trivial $\zeta$ zeros, that are poles of $\frac{\zeta'}{\zeta}$
I now want to use this to approximate $$\frac{1}{2\pi i}\int_{-U-iT}^{-U+iT}\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\mathrm{ds}$$ with $U\in\mathbb{N}$ being odd.
This obviously gives
\begin{align*} \Bigg\vert\frac{1}{2\pi i}\int_{-U-iT}^{-U+iT}\frac{\zeta'}{\zeta}(s)\frac{x^s}{s}\mathrm{ds} \Bigg\vert &\ll \int_{-T}^{T}\Bigg\vert\frac{\zeta'}{\zeta}(-U+it)\Bigg\vert\frac{x^{-U}}{\vert -U+it\vert}\mathrm{dt}\\ &\ll \frac{x^{-U}}{U}\int_{-T}^{T}\Bigg\vert\frac{\zeta'}{\zeta}(-U+it)\Bigg\vert\mathrm{dt} \end{align*}
and the author just writes, that due to the approximation at the begining, this all is just $O(\frac{\log U}{U}x^{-U}T)$. But I do not understand where the $\log U$ comes from?
I know that If I use aboves approximation, I can somehow pull $\frac{\zeta'}{\zeta}$ outside the integral, and evaluating the empty integral and get $\int_{-T}^{T}\mathrm{dt}=O(T)$.
But how do I get $\log U$? I mean I should only get $\log\vert -U+it\vert$ and then I am unable to pull this outside the integral. And since the logarithm is monotonic I dont see how I could justify to just drop the $it$ right there?