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Question Number 61240 by Tawa1 last updated on 30/May/19

∫ ((x^(2 ) − 4)/((x^2 + 4)^2 )) dx

$$\int\:\frac{\mathrm{x}^{\mathrm{2}\:} −\:\mathrm{4}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$

Commented by maxmathsup by imad last updated on 31/May/19

let A =∫ ((x^2 −4)/((x^2 +4)^2 ))dx ⇒ A = ∫ ((x^2 +4−8)/((x^2 +4)^2 ))dx =∫ (dx/(x^2 +4))−8 ∫ (dx/((x^2 +1)^2 )) ∫ (dx/(x^2 +4)) =_(x =2t) ∫ ((2dt)/(4(1+t^2 ))) =(1/2) arctan(t) +c_0 =(1/2) arctan((x/2))+c_0 ∫ (dx/((x^2 +1)^2 )) =_(x =tanθ) ∫ (((1+tan^2 θ))/((1+tan^2 θ)^2 ))dθ =∫ (dθ/(1+tan^2 θ)) =∫ cos^2 θ dθ =(1/2) ∫ (1+cos(2θ))dθ =(1/2)θ +(1/4)sin(2θ) +c_1 =(θ/2) +(1/2) sinθcosθ =(1/2) arctan(x)+(1/2) (x/(√(1+x^2 ))) (1/(√(1+x^2 ))) =(1/2) arctan(x)+(x/(2(1+x^2 ))) +c_1 ⇒ A =(1/2) arctan((x/2))−4arctan(x)−((4x)/(1+x^2 )) +C .

$${let}\:{A}\:=\int\:\:\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\:{A}\:=\:\int\:\frac{{x}^{\mathrm{2}} \:+\mathrm{4}−\mathrm{8}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}−\mathrm{8}\:\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=_{{x}\:=\mathrm{2}{t}} \:\:\:\:\int\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{4}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left({t}\right)\:+{c}_{\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+{c}_{\mathrm{0}} \\ $$$$\:\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=_{{x}\:={tan}\theta} \:\:\:\:\:\int\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta\:=\int\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta} \\ $$$$=\int\:\:\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\theta\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right)\:+{c}_{\mathrm{1}} \:=\frac{\theta}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\theta{cos}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left({x}\right)+\frac{{x}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:+{c}_{\mathrm{1}} \:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{4}{arctan}\left({x}\right)−\frac{\mathrm{4}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+{C}\:. \\ $$

Commented by Tawa1 last updated on 01/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by maxmathsup by imad last updated on 01/Jun/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Answered by MJS last updated on 30/May/19

in such cases I try the following: δ(x)=(d/dx)[((f(x))/(x^2 +4))]=((f′(x)(x^2 +4)−2f(x)x)/((x^2 +4)^2 )) if f(x)=c ⇒ δ(x)=−((2cx)/((x^2 +4)^2 )) ⇒ we need another x if f(x)=cx ⇒ δ(x)=((−c(x^2 −4))/((x^2 +4)^2 )) ⇒ c=−1 ⇒ f(x)=−x ⇒ ∫((x^2 −4)/((x^2 +4)^2 ))dx=−(x/(x^2 +4))

$$\mathrm{in}\:\mathrm{such}\:\mathrm{cases}\:\mathrm{I}\:\mathrm{try}\:\mathrm{the}\:\mathrm{following}: \\ $$$$\delta\left({x}\right)=\frac{{d}}{{dx}}\left[\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}}\right]=\frac{{f}'\left({x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)−\mathrm{2}{f}\left({x}\right){x}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\mathrm{if}\:{f}\left({x}\right)={c}\:\Rightarrow\:\delta\left({x}\right)=−\frac{\mathrm{2}{cx}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{need}\:\mathrm{another}\:{x} \\ $$$$\mathrm{if}\:{f}\left({x}\right)={cx}\:\Rightarrow\:\delta\left({x}\right)=\frac{−{c}\left({x}^{\mathrm{2}} −\mathrm{4}\right)}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{c}=−\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)=−{x}\:\Rightarrow \\ $$$$\int\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$

Commented by Tawa1 last updated on 01/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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