= Solution
Choose the branch $\operatorname{Log}z=\log|z|+i\arg z$ with $-\pi/2<\arg z<3\pi/2$. It is <holomorphic> off the downward imaginary ray and takes values $\log x$ on the positive real axis and $\log x+i\pi$ at $-x$ for $x>0$.
Take $R>2$ and $0<\varepsilon<2$. The positively oriented <contour> consists of the real segments $[-R,-\varepsilon]$ and $[\varepsilon,R]$, the small upper semicircle from $-\varepsilon$ to $\varepsilon$ traversed clockwise, and the large upper semicircle from $R$ to $-R$ traversed counterclockwise. For $F(z)=\operatorname{Log}z/(4+z^2)$, the only enclosed <pole> is $2i$, with
$$
\operatorname{Res}_{z=2i}F=\frac{\log2+i\pi/2}{4i}.
$$
The integral over the large arc is $O(\log R/R)$, since its length is $\pi R$ and $|4+z^2|\ge R^2-4$. The small arc contributes $O(\varepsilon|\log\varepsilon|)$, since its length is $\pi\varepsilon$ and $|4+z^2|\ge4-\varepsilon^2$. Both tend to zero.
Let $J=\int_0^\infty\log x/(4+x^2)\,dx$; it converges because $|\log x|$ is integrable near zero and $\log x/x^2$ is integrable at infinity. The two real segments, after substituting $z=-x$ on the negative one, tend to
$$
2J+i\pi\int_0^\infty\frac{dx}{4+x^2}=2J+\frac{i\pi^2}{4}.
$$
The <residue theorem> therefore gives
$$
2J+\frac{i\pi^2}{4}
=2\pi i\frac{\log2+i\pi/2}{4i}
=\frac\pi2\log2+\frac{i\pi^2}{4}.
$$
Taking real parts yields
$$
\boxed{J=\frac\pi4\log2.}
$$
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