The Riemann-von Mangoldt formula for the number of non-trivial zeros of $\zeta(s)$ with imaginary part between $0$ and $T$ is given by $$ N(T) = \#\{\rho = \beta + i\gamma \in \mathbb{C} : \zeta(\rho) = 0 \text{ and } 0 < \beta < 1, 0 \leq \gamma \leq T\} = \frac{T}{2\pi} \log \frac{T}{2 \pi} - \frac{T}{2 \pi} + O(\log T). \label{riemann mangoldt formula} \tag{1} $$
From Davenport's book, I know that \begin{equation} \sum_{\gamma} \frac{1}{1 + (t - \gamma)^2} = O(\log t), \label{my eqn} \tag{2} \end{equation} where $\gamma$ runs over the non-trivial zeros of $\zeta(s)$. I want to obtain an estimate for the sum $$ \sum_{\gamma \not\in [0, T]} \frac{1}{1 + (t - \gamma)^2} $$ when $t \in [0, T]$ using \eqref{riemann mangoldt formula} and \eqref{my eqn}. I tried using a modified version of Abel's summation to obtain $$ \sum_{\gamma > T} \frac{1}{1 + (t - \gamma)^2} = - \frac{N(T)}{1 + (t - T)^2} - 2 \int_{T}^{\infty} \frac{N(x)(t-x)}{(1 + (t-x)^2)^2}\, dx $$ and $$ \sum_{\gamma < 0} \frac{1}{1 + (t - \gamma)^2} = 2 \int_{0}^{\infty} \frac{N(x)(t + x)}{(1 + (t + x)^2)^2} \, dx, $$ but I seem to not being able to cleverly solve or manipulate the integral to get estimate of the form $$ \sum_{\gamma \not\in [0, T]} \frac{1}{1 + (t - \gamma)^2} \ll \left( \frac{1}{t+1} + \frac{1}{T-t+1} \right) \log T $$ for $t \in [0, T]$. Any hint or suggestion will be greatly appreciated.