Question Number 78134 by msup trace by abdo last updated on 14/Jan/20
$${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{3}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +\mathrm{4}}{dx} \\ $$
Commented by mathmax by abdo last updated on 14/Jan/20
$${let}\:{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{3}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\:{and}\:{W}\left({z}\right)=\frac{{arctan}\left(\mathrm{3}{z}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${W}\left({z}\right)=\frac{{arctan}\left(\mathrm{3}{z}^{\mathrm{2}} \right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:\Rightarrow\int_{−\infty} ^{+\infty\:} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},\mathrm{2}{i}\right) \\ $$$$=\mathrm{2}{i}\pi×\frac{\mid{arctan}\left(\mathrm{3}\left(\mathrm{2}{i}\right)^{\mathrm{2}} \right)\mid}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\mid{arctan}\left(−\mathrm{12}\right)\mid\:=\frac{\pi}{\mathrm{2}}{arctan}\left(\mathrm{12}\right)\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{2}}{arctan}\left(\mathrm{12}\right) \\ $$