Interesting Limit (29)

Substitute $x:=a y,\: dx=a\:dy$ : $F(a):=\frac{1}{a}\int_0^a\arctg(x)\arctg(a-x)\:dx=\cancel{\frac{1}{a}}\int_0^1\arctg(ay)\arctg(a(1-y))\cdot \cancel{a}\:dy=:\int_0^1 f_a(y)\:dy$ We note $|f_a(y)|\leq\frac{\pi^2}{4}=:f(y)$ integrable and $f_a(y)\rightarrow f(y)$ almost everywhere on $[0;\:1]$ (only borders are exceptions). By dominated convergence , we have $\begin{aligned} \lim_{a\rightarrow\infty} F(a)&=\lim_{a\rightarrow\infty}\int_0^1 f_a(x)\:dx=\int_0^1 f(x)\:dx=\frac{\pi^2}{4}&&&\Rightarrow &&&& A+B&=2+4=\boxed{6} \end{aligned}$