I am trying to compute the real integral $$\int_{0}^{\infty} \frac{\ln(x)}{(x^2+1)^2} \ dz,$$ by computing the contour integral$$\int_{0}^{\infty}\frac{\text{Log}(z)}{(z^2+1)^2} \ dz, \ \ \ \text{where $\frac{-\pi}{2}<\text{Arg}\leq\frac{3\pi}{2}$}$$
I have computed both the lines from $-R$ to $-r$ and $r$ to $R$ shown on the diagram below. However, I am having trouble computing the circular regions via the ML Lemma ($r<1<R$).
For the half-circle with radius $R$, I have tried: \begin{align} \text{let} \ z&=Re^{it} \ \ t\in [0,\pi] \\ \implies |z|&=R \\ \text{consider} \ |z^2+1|&\geq ||z^2|-|-1|| \\ \frac{1}{|(z^2+1)^2|}&\leq\frac{1}{(R^2-1)^2} \\ \left|\frac{\text{Log}(z)}{(z^2+1)^2}\right|&\leq\frac{\text{Log}(z)}{(R^2-1)^2} \end{align}
But does $$\frac{\text{Log}(z)}{(R^2-1)^2} \times R\pi\rightarrow 0\ \text{as} \ R\rightarrow\infty?$$
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update
I have solved the integral, with a lot of help from the comments section; thanks! But my proof requires that the inner cirlce with radius $r$ is $0$: $$\left|\frac{\text{Log}(z)}{(z^2+1)^2}\right|\leq \frac{\ln(r)-i\pi}{(1-r^2)^2}\rightarrow 0 \ \text{as} \ r\rightarrow 0.$$ I have used the ML Lemma, but I do not know how to evaluate this limit (it does not appear to be in indeterminate form, hence L'Hopital's rule does not apply). How do I evaluate the limit (or show the contour of the inner circle is $0$ using the ML lemma)?
An alternative approach: for any $s\in(-1,3)$, by the substitution $x\mapsto\tan\theta$ and Euler's Beta function we have
$$ f(s)=\int_{0}^{+\infty}\frac{x^s}{(x^2+1)^2}\,dx = \frac{\pi(1-s)}{4\cos\left(\frac{\pi s}{2}\right)}=\frac{\pi}{4}(1-s)(1+O(s^2))$$ hence by differentiating both sides with respect to $s$ $$ f'(s) = \int_{0}^{+\infty}\frac{x^s\log x}{(1+x^2)^2}\,dx = -\frac{\pi}{8\cos\left(\frac{\pi s}{2}\right)}\left[2-\pi\tan\left(\tfrac{\pi s}{2}\right)+\pi s\tan\left(\tfrac{\pi s}{2}\right)\right] $$ and the Maclaurin series of the RHS is $\color{red}{-\frac{\pi}{4}}+\frac{\pi^3}{16}s+O(s^2)$.
Even shorter: by the substitution $x\mapsto\frac{1}{x}$ we have $f(s)=f(2-s)$, and by integration by parts $\frac{f(s)}{1-s}$ is an even function. In particular $f'(0)$ (which is the wanted integral) simply equals $$-f(0)=-\int_{0}^{+\infty}\frac{dx}{(1+x^2)^2}=-\frac{\pi}{4}.$$