Let $a,b\in(0,1)$.
In the course of solving Evaluating a double integral of a complicated rational function we came across the following definite integrals: \begin{eqnarray} {\mathfrak J}_1(a,b) &:=& \int\limits_{-\infty}^\infty \frac{\arctan(\frac{b}{x})}{(x+a/2)^2+(1-b)^2/4} dx\\ {\mathfrak J}_2(a,b) &:=& \int\limits_{-\infty}^\infty \frac{\arctan(\frac{1}{x+a})}{(x+a/2)^2+(1-b)^2/4} dx \end{eqnarray}
Since $\arctan(x)=\imath/2(\log(1-\imath x)-\log(1+\imath x))$ and since the inverse of the denominator can always be decomposed into partial fractions the anti-derivatives of both integrands above can be found and are expressed through logs and di-logarithms only. Then the values of those anti-derivatives at positive and negative infinities can be taken and the integrals above evaluated. We have therefore carried out those calculations and obtained the following results: \begin{equation} {\mathfrak J}_1(a,b)=-\frac{2 \pi}{1-b} \arctan(\frac{2 a b}{1+a^2-b^2}) \end{equation} The second integral is more complicated since, as it turns out, the anti-derivative has two discontinuity points in the real axis at $x\in \left\{-a,-a b/(1+b)\right\}$. We have: \begin{eqnarray} &&{\mathfrak J}_2(a,b)=\frac{1}{2(1-b)} \sum\limits_{\xi=\pm} \sum\limits_{\eta=\pm} \xi \eta\\ && \left( \right.\\ &&\left. (g_+^{(\eta,\xi)}-g^{(\eta,\xi)}(-\frac{a b}{1-b}+\epsilon))+\right.\\ &&\left. (g^{(\eta,\xi)}(-\frac{a b}{1-b}-\epsilon)-g^{(\eta,\xi)}(-a+\epsilon))+\right.\\ &&\left. (g^{(\eta,\xi)}(-a-\epsilon)-g_-^{(\eta,\xi)}\right.)\\ &&\left. \right) \end{eqnarray}
where \begin{eqnarray} &&g^{(\eta,\xi)}(x):= \frac{1}{2} \log ^2\left(-\frac{2 (a+x)}{a+i b \eta -i \eta }\right)-\frac{1}{2} \log ^2\left(-\frac{2 (a+x+i \xi )}{2 (a+i \xi )-a+i b \eta -i \eta }\right)+\\ &&\log (a+x+i \xi ) \log \left(1-\frac{2 (a+x+i \xi )}{2 (a+i \xi )-a+i b \eta -i \eta }\right)-\log (a+x) \log \left(1-\frac{2 (a+x)}{a+i b \eta -i \eta }\right)+\\ &&\text{Li}_2\left(\frac{a+i b \eta -i \eta }{2 (a+x)}\right)-\text{Li}_2\left(\frac{-a+i b \eta -i \eta +2 (a+i \xi )}{2 (a+x+i \xi )}\right)+\\ &&2\pi \imath 1_{\xi=-1} 1_{x\le -a} \log(a-\imath (-1+b) \eta +2 x) \end{eqnarray} and \begin{eqnarray} &&g_+^{(\eta,\xi)} = \frac{1}{2} \left( \log(\frac{-2}{a+\imath(-1+b)\eta})^2 - \log(\frac{-2}{a+\imath((-1+b)\eta+2\xi)})^2\right)\\ &&g_-^{(\eta,\xi)} = \frac{1}{2} (-\log (a-i (1-b) \eta )-\log (a-i (-b \eta +\eta +2))+2 \log (2))\cdot \\ && (\log (a-i (-b \eta +\eta +2))-\log (a-i (1-b) \eta ))+\\ &&\left\{ \begin{array}{r} i \pi (\log (a-i (1-b) \eta )-\log (a+i (b \eta -\eta +2))) & \mbox{if $\xi=1$} \\ -2 \pi ^2 \eta +i \pi (\log (a-i (1-b \eta ))+\log (a-i (-b \eta +2 \eta +1))) & \mbox{if $\xi=-1$} \end{array} \right. \end{eqnarray} Now we expand the complex quantities and simplify the result and again we get neat closed form results. We have: \begin{eqnarray} \sum\limits_{\xi=\pm} \sum\limits_{\eta=\pm} \eta \xi(g^{(\eta,\xi)}_+-g^{(\eta,\xi)}_-) &=& -4 \pi\left(\pi+\arctan(\frac{2 a(-1+b)}{a^2-(-3+b)(1+b)})\right) \\ \sum\limits_{\xi=\pm} \sum\limits_{\eta=\pm} \eta \xi( g^{(\eta,\xi)}(-\frac{a b}{1+b}-\epsilon)-g^{(\eta,\xi)}(-\frac{a b}{1+b}+\epsilon)) &=& 4\pi \arctan(\frac{1+b}{a})\\ \sum\limits_{\xi=\pm} \sum\limits_{\eta=\pm} \eta \xi( g^{(\eta,\xi)}(-a-\epsilon)-g^{(\eta,\xi)}(-a+\epsilon)) &=& 4 \pi\left(\pi - \arctan(\frac{1-b}{a}) \right) \end{eqnarray} Putting everything together we have: \begin{equation} {\mathfrak J}_2(a,b)= \frac{2 \pi}{(1-b)} \arctan\left( \frac{2 a}{3+a^2-4 b+b^2}\right) \end{equation} Now, the obvious question would be is it possible to derive those results using some other method, for example using complex analysis.