Let $\gamma$ be the upper half the circunference $C(0,R)$ with counter-clockwise orientation
How to prove that
$\displaystyle\int_{\gamma}\frac{(\ln z)^2}{z^2 +1} \rightarrow 0$ as $R \rightarrow \infty$
by my calculations this was limited by a constant that does't depends on my $R$
$\displaystyle \frac{(\ln z)^2}{z^2 +1} = \frac{(\ln|z|+iargz)^2}{z^2 +1} \leq \frac{ |z|^2A}{z^2 +1}$ $A$ is something limited as $R \rightarrow \infty$ and in module this quotient tends to a constant $C\neq 0$.
If $z=Re^{i\theta}$, $R>1$ and $\theta\in(0,\pi)$, $$\left\|\frac{(\log z)^2}{z^2+1}\right\|\leq \frac{\left\|\log z\right\|^2}{R^2-1}=\frac{\log^2R+\theta^2}{R^2-1}$$ so $$ \int_\gamma\frac{\log^2 z}{z^2+1}\,dz \ll \frac{\log^2 R}{R} $$ as $R\to +\infty$.