Contour integral for finding $\int_0^\infty \frac{\ln x}{(x+a)^2+b^2} \, dx$

Note: my solution works only for $a>0,\,b\neq0$. Consider the function $$f\left(z\right)=\frac{\log^{2}\left(z\right)}{\left(z-a\right)^{2}+b^{2}} $$ and the branch of the logarithm corresponding to $-\pi<\arg\left(z\right)\leq\pi.$ Take the keyhole contour and define $\Gamma$ and $\gamma$ respectively the large circumference of radius $R$ and the small circumference of radius $\rho$.

We note that $$\left|\int_{\Gamma}f\left(z\right)dz\right|\leq\frac{\log^{2}\left(R\right)+\pi^{2}}{\left(R-\left|a\right|\right)^{2}-b^{2}}2\pi R\underset{R\rightarrow\infty}{\rightarrow}0 $$ and $$\left|\int_{\gamma}f\left(z\right)dz\right|\leq\frac{\log^{2}\left(\rho\right)+\pi^{2}}{\left|\left(\rho-\left|a\right|\right)^{2}-b^{2}\right| }2\pi\rho\underset{\rho\rightarrow0}{\rightarrow}0. $$ So we have $$2\pi i\left(\textrm{Res}_{z=a+ib}\frac{\log^{2}\left(z\right)}{\left(z-a\right)^{2}+b^{2}}+\textrm{Res}_{z=a-ib}\frac{\log^{2}\left(z\right)}{\left(z-a\right)^{2}+b^{2}}\right) $$ $$=\int_{0}^{\infty}\frac{\log^{2}\left(-x+i\epsilon\right)}{\left(-x+i\epsilon-a\right)^{2}+b^{2}}dx-\int_{0}^{\infty}\frac{\log^{2}\left(-x-i\epsilon\right)}{\left(-x-i\epsilon-a\right)^{2}+b^{2}}dx $$ $$\underset{\epsilon\rightarrow0}{\rightarrow}\int_{0}^{\infty}\frac{\left(\log\left(x\right)+i\pi\right)^{2}-\left(\log\left(x\right)-i\pi\right)^{2}}{\left(x+a\right)^{2}+b^{2}}dx $$ $$=4\pi i\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+a\right)^{2}+b^{2}}dx $$ and for the other part we have $$\textrm{Res}_{z=a+ib}\frac{\log^{2}\left(z\right)}{\left(z-a\right)^{2}+b^{2}}+\textrm{Res}_{z=a-ib}\frac{\log^{2}\left(z\right)}{\left(z-a\right)^{2}+b^{2}}=\frac{\log^{2}\left(a+ib\right)-\log^{2}\left(a-ib\right)}{2ib}. $$ Now if we assume that $a>0 $ we have $$\frac{\left(\log\left(\sqrt{a^{2}+b^{2}}\right)+i\arg\left(a+ib\right)\right)^{2}-\left(\log\left(\sqrt{a^{2}+b^{2}}\right)+i\arg\left(a-ib\right)\right)^{2}}{2ib} \tag{1}$$ $$=\frac{2\log\left(\sqrt{a^{2}+b^{2}}\right)\arctan\left(\frac{b}{a}\right)}{b}. $$ Hence

$$\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+a\right)^{2}+b^{2}}dx=\frac{\log\left(\sqrt{a^{2}+b^{2}}\right)\arctan\left(\frac{b}{a}\right)}{b}. $$

Addendum: the identity holds only if $a>0 $. If we assume $a<0$ from $(1)$ we have

$$\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+a\right)^{2}+b^{2}}dx=\frac{\log\left(\sqrt{a^{2}+b^{2}}\right)\left(\pi-\arctan\left(\frac{b}{a}\right)\right)}{b}. $$

An alternative, real-analytic solution by symmetry only.

$$ \int_{0}^{+\infty}\frac{\log x}{x^2+2ax+(a^2+b^2)}\stackrel{x\mapsto u\sqrt{a^2+b^2}}{=}\frac{1}{\sqrt{a^2+b^2}}\int_{0}^{+\infty}\frac{\log u+\log\sqrt{a^2+b^2}}{u^2+\frac{2a}{\sqrt{a^2+b^2}}u+1}\,du $$ and since the polynomial $p(u)=u^2+\frac{2a}{\sqrt{a^2+b^2}}u+1$ is quadratic and palindromic,
$\int_{0}^{+\infty}\frac{\log u}{p(u)}\,du=0$ by the substitution $u\mapsto \frac{1}{u}$. This leaves us with the elementary integral $$ \int_{0}^{+\infty}\frac{du}{u^2+2Du+1}=\int_{D}^{+\infty}\frac{du}{u^2+(1-D^2)}=\frac{1}{\sqrt{1-D^2}}\,\left(\frac{\pi}{2}-\arctan\frac{D}{\sqrt{1-D^2}}\right) $$ for any $D\in(-1,1)$. By letting $D=\frac{a}{\sqrt{a^2+b^2}}$ we get $$\int_{0}^{+\infty}\frac{\log x}{(x+a)^2+b^2}=\frac{\log\sqrt{a^2+b^2}}{|b|}\left[\frac{\pi}{2}-\arctan\left(\frac{a}{|b|}\right)\right] $$ for any $b\neq 0$. If $b=0$ the LHS is convergent only for $a>0$, and in such a case it equals $\frac{\log a}{a}$.