The function $ln(1+x^2)$ has two branch points at $\pm i$, then we cannot choose the contour you propose, as noted in the comment by @Sangchul Lee. Also, the point $x=0$ is not special for the function, the contribution of $\gamma_r$ is useless too. Instead, we can consider a branch cut on the vertical segment between $i$ and $-i$.
It is convenient to decompose
$$ f(x)=\frac{1}{2}\frac{\ln(1+x^2)}{1-ix}+\frac{1}{2}\frac{\ln(1+x^2)}{1+ix}$$
By enforcing the substitution $y=-x$ in the second integral, one has
\begin{align}
I&=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^2)}{1-ix}\,dx+\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^2)}{1+ix}\,dx\\
&=\int_{-\infty}^\infty\frac{\ln(1+x^2)}{1-ix}\,dx
\end{align}
We define a contour along the real axis which avoids the branch cut by turning around $x=i$ and which is closed by the large upper semi-circle $C_R$. From the residue theorem, it holds
\begin{equation}
0=\left[\int_{C_R}+\int_{-\infty}^{-\varepsilon}+J+\int_{\varepsilon}^{\infty}\right] f^-(x)\,dx
\end{equation}
where $f^-(x)=\left( 1-ix \right)^{-1}\ln\left( 1+x^2 \right)$ and $J$ is the contribution of the path which turns around the branch cut. It can be shown that the large semi-circle contribution vanishes as $R\to\infty$, then
\begin{equation}
I=\int_{-\infty}^\infty f^-(x)\,dx=-J
\end{equation}
and
$$J=\left[\int_{-\varepsilon}^{i-\varepsilon} +\int_{\gamma_i} +\int^{\varepsilon}_{i+\varepsilon} \right] f^-(x)\,dx$$
$\int_{\gamma_i}$ is the contribution of a small semi-circle around $x=i$, it can be shown to vanish when $\varepsilon\to 0$. On the left and right sides of the vertical contribution, with $x=-\varepsilon+ iy$, and passing to the limit $\varepsilon\to 0$
\begin{align}
\int_{-\varepsilon}^{i-\varepsilon}f^-(x)\,dx&=i\int_0^1\frac{\ln(1-y^2)+2i\pi}{1+y}\,dy\\
\int^{\varepsilon}_{i+\varepsilon}f^-(x)\,dx&=i\int_1^0\frac{\ln(1-y^2)}{1+y}\,dy
\end{align}
Adding both terms, the logarithmic contributions cancel each other and we obtain
\begin{equation}
J=-2\pi\ln2
\end{equation}
Finally,
\begin{equation}
I=2\pi\ln2
\end{equation}
In this method, like in the other given answers, a decomposition is used in order to separate the poles and the branch points near the contour.