= Solution
The <residue theorem> states that a positively oriented simple closed contour enclosing isolated poles and none on the contour has integral $2\pi i$ times the sum of their <residues>, provided the function is holomorphic elsewhere inside. Near a simple zero of $\psi$, $\psi(z)=\psi'(a)(z-a)+O((z-a)^2)$. Hence
$$
\boxed{\operatorname{Res}_{z=a}\frac{\phi(z)}{\psi(z)}=\lim_{z\to a}\frac{(z-a)\phi(z)}{\psi(z)}=\frac{\phi(a)}{\psi'(a)}.}
$$
For $\cot z=\cos z/\sin z$, this gives $\boxed{\operatorname{Res}_{z=k\pi}\cot z=1}$ for every integer $k$.
To evaluate the real integral, integrate $F(z)=\operatorname{Log}z/(z^2+a^2)$ around an upper semicircle, with a small upper semicircular indentation excluding zero. Use the branch real on the positive axis and with argument between zero and $\pi$ in the upper half-plane. The large arc tends to zero as $O(\ln R/R)$ and the small arc as $O(\epsilon|\ln\epsilon|)$. The only enclosed pole is $ia$, whose contribution is
$$
2\pi i\operatorname{Res}_{z=ia}F=\frac\pi a\left(\ln a+\frac{i\pi}{2}\right).
$$
On the negative real axis the upper boundary value is $\operatorname{Log}(-x)=\ln x+i\pi$. The two real segments therefore contribute
$$
2I+i\pi\int_0^\infty\frac{dx}{x^2+a^2}=2I+\frac{i\pi^2}{2a}.
$$
Comparing real parts yields
$$
\boxed{I=\int_0^\infty\frac{\ln x}{x^2+a^2}dx=\frac{\pi\ln a}{2a},\qquad I|_{a=1}=0.}
$$
The integral is absolutely convergent at both zero and infinity, so no principal-value interpretation is needed.
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