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Question Number 93473 by mashallah last updated on 13/May/20

∫1/(1+x^2 )^2

$$\int\mathrm{1}/\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$

Commented by abdomathmax last updated on 15/May/20

A =∫  (dx/((1+x^2 )^2 ))  we do the changement x =tant ⇒  A  =∫   (((1+tan^2 t)dt)/((1+tan^2 t)^2 )) =∫cos^2 t dt =(1/2)∫(1+cos(2t))dt  =(t/2) +(1/4)sin(2t) +C  =((arctanx)/2) +(1/4)×((tant)/(1+tan^2 t)) +C  =((arctanx)/2) +(1/4)(x/(1+x^2 )) +C

$${A}\:=\int\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{we}\:{do}\:{the}\:{changement}\:{x}\:={tant}\:\Rightarrow \\ $$$${A}\:\:=\int\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:=\int{cos}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{t}\right)\:+{C} \\ $$$$=\frac{{arctanx}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}×\frac{{tant}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:+{C} \\ $$$$=\frac{{arctanx}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+{C} \\ $$$$ \\ $$

Answered by Rio Michael last updated on 13/May/20

∫ (1/((1 + x^2 )^2 )) dx     changement  x = tan θ ⇒ (dx/dθ) = sec^2 θ  ⇒ ∫(1/((1+x^2 )^2 )) dx = ∫ (1/((1+tan^2 θ)^2 )) sec^2 θ dθ = ∫(1/(sec^2 θ)) dθ  ⇒ ∫ cos^2 θ dθ = (1/2) ∫ (1 +cos2θ) dθ = (1/2) ( θ + ((sin 2θ)/2)) + C

$$\int\:\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:\:\: \\ $$$$\mathrm{changement}\:\:{x}\:=\:\mathrm{tan}\:\theta\:\Rightarrow\:\frac{{dx}}{{d}\theta}\:=\:\mathrm{sec}^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=\:\int\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta\:=\:\int\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$\Rightarrow\:\int\:\mathrm{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\mathrm{1}\:+\mathrm{cos2}\theta\right)\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\theta\:+\:\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}\right)\:+\:{C} \\ $$$$ \\ $$

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