 Solution

Use the principal branch of the complex logarithm for $\mathrm{Log}(z+i)$. Its branch cut is the ray $z=-i-s$, $s\ge 0$, lying below the real axis. Thus it is analytic in the upper half-plane, where the only pole of the integrand is $z=i$.

Close the real interval by a large upper semicircle. The estimation lemma bounds the arc integral by $O(\ln R/R)$, so it vanishes. The simple-pole residue isThe residue theorem then givesFor real $x$, $\mathrm{Re}\mathrm{Log}(x+i)=\frac{1}{2}\ln (x^2+1)$. Taking real parts and then using evenness of the real integrand yieldsBoth endpoints converge: the integrand is $O(x^2)$ near zero and $O(\ln x/x^2)$ at infinity. This also agrees with the logarithmic parameter integral with Cauchy weight at parameter one.