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Question Number 52649 by Tawa1 last updated on 10/Jan/19

∫ ((4x^2 + 3)/((x^2 + x + 1)^2 )) dx

$$\int\:\frac{\mathrm{4x}^{\mathrm{2}} \:+\:\mathrm{3}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$

Commented by maxmathsup by imad last updated on 10/Jan/19

et I =∫ ((4x^2 +3)/((x^2 +x+1)^2 ))dx ⇒ I =∫ ((4(x^2 +x+1)−4x−1)/((x^2 +x+1)^2 ))dx =4 ∫ (dx/(x^2 +x +1)) −∫ ((4x+1)/((x^2 +x+1)^2 ))dx =4H −K H =∫ (dx/((x+(1/2))^2 +(3/4))) =_(x+(1/2)=((√3)/2)tanθ) ∫ (1/((3/4)(1+tan^2 θ))) ((√3)/2) (1+tan^2 θ)dθ =((√3)/2) (4/3) ∫ dθ =(2/(√3)) θ +c_1 =(2/(√3)) arctan(((2x+1)/(√3))) +c_1 we have ((1/(x^2 +x+1)))^′ =−((2x+1)/((x^2 +x +1))) ⇒ ∫ ((4x+1)/((x^2 +x+1)^2 )) =2 ∫ ((2x+(1/2))/((x^2 +x+1)^2 ))dx =2 ∫ ((2x+1−(1/2))/((x^2 +x+1)^2 ))dx =2{−(1/(x^2 +x+1))−(1/2) ∫ (dx/((x^2 +x+1)^2 ))} =((−2)/(x^2 +x+1)) −∫ (dx/((x^2 +x+1)^2 )) but ∫ (dx/((x^2 +x+1)^2 )) =∫ (dx/(((x+(1/2))^2 +(3/4))^2 )) =_(x+(1/2)=((√3)/2)tanθ) ((16)/9)((√3)/2) ∫ (1/((1+tan^2 θ)^2 )) (1+tan^2 θ)dθ =((8(√3))/9) ∫ (dθ/(1+tan^2 θ)) =((8(√3))/9) ∫ ((1+cos(2θ))/2)dθ =((4(√3))/9) ∫ (1+cos(2θ))dθ =((4(√3))/9) θ +((4(√3))/(18)) sin(2θ) =((4(√3))/9) arctan(((2x+1)/(√3))) +((2(√3))/9) sin(2arctan(((2x+1)/(√3))))+c_2 ⇒ I =(8/(√3)) arctan(((2x+1)/(√3))) +(2/(x^2 +x+1)) +((4(√3))/9) arctan(((2x+1)/(√3)))+((2(√3))/9) sin(2arctan(((2x+1)/(√3)))) +C

$${et}\:{I}\:=\int\:\:\frac{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{3}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\:{I}\:=\int\:\frac{\mathrm{4}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)−\mathrm{4}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}}\:−\int\:\:\:\frac{\mathrm{4}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:=\mathrm{4}{H}\:−{K} \\ $$$${H}\:=\int\:\:\:\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}\theta} \:\:\:\int\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{3}}\:\int\:\:{d}\theta\:=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\theta\:+{c}_{\mathrm{1}} =\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+{c}_{\mathrm{1}} \\ $$$${we}\:{have}\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\right)^{'} =−\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\int\:\:\:\frac{\mathrm{4}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\mathrm{2}\:\int\:\frac{\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:=\mathrm{2}\:\int\:\:\frac{\mathrm{2}{x}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{2}\left\{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{−\mathrm{2}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:−\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\int\:\:\frac{{dx}}{\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }\:=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}\theta} \:\:\frac{\mathrm{16}}{\mathrm{9}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta\:=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int\:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\theta\:\:\:+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{18}}\:{sin}\left(\mathrm{2}\theta\right)\:=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\:{sin}\left(\mathrm{2}{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right)+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{8}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+\frac{\mathrm{2}}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}\:+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\:{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\:{sin}\left(\mathrm{2}{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right)\:+{C} \\ $$

Commented by Tawa1 last updated on 10/Jan/19

God bless you sir

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

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