Question Number 93473 by mashallah last updated on 13/May/20
$$\int\mathrm{1}/\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$
Commented by abdomathmax last updated on 15/May/20
$${A}\:=\int\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{we}\:{do}\:{the}\:{changement}\:{x}\:={tant}\:\Rightarrow \\ $$$${A}\:\:=\int\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:=\int{cos}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{t}\right)\:+{C} \\ $$$$=\frac{{arctanx}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}×\frac{{tant}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:+{C} \\ $$$$=\frac{{arctanx}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+{C} \\ $$$$ \\ $$
Answered by Rio Michael last updated on 13/May/20
$$\int\:\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:\:\: \\ $$$$\mathrm{changement}\:\:{x}\:=\:\mathrm{tan}\:\theta\:\Rightarrow\:\frac{{dx}}{{d}\theta}\:=\:\mathrm{sec}^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=\:\int\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta\:=\:\int\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$\Rightarrow\:\int\:\mathrm{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\mathrm{1}\:+\mathrm{cos2}\theta\right)\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\theta\:+\:\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}\right)\:+\:{C} \\ $$$$ \\ $$