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x-6-1-x-2-1-

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Question Number 29517 by yesaditya22@gmail.com last updated on 09/Feb/18
∫x^6 −1/x^2 +1
$$\int\mathrm{x}^{\mathrm{6}} −\mathrm{1}/\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$
Commented by abdo imad last updated on 09/Feb/18
let put I= ∫ ((x^6 −1)/(x^2 +1))dx = ∫ ((x^6 +x^4 −x^4 −1)/(x^2 +1))dx = ∫ x^4 dx −∫ ((1+x^4 )/(1+x^2 ))dx = ∫x^4 dx −∫(dx/(1+x^2 )) −∫ (x^4 /(1+x^2 ))dx =(x^5 /5) −arctanx −∫ (x^4 /(1+x^2 ))dx the ch.x=tanθ give ∫ (x^4 /(1+x^2 ))dx= ∫ ((tan^4 θ)/(1+tan^2 θ))(1+tan^2 θ)dθ= ∫ tan^4 θ dθ but = ∫((1−cos(2θ))^2 )/((1+cos(2θ))^2 ))dθ =∫ ((1−2cos(2θ) +((1+cos(4θ))/2))/(1+2cos(2θ)+((1+cos(4θ))/2)))dθ =∫ ((2−4cos(2θ) +1+cos(4θ))/(2 +4cos(2θ) +1 +cos(4θ)))dθ =∫ ((3 −4cos(2θ) +cos(4θ))/(3 +4cos(2θ) +cos(4θ)))dθ this integral is calculable if the lmits are given...
$${let}\:{put}\:{I}=\:\int\:\frac{{x}^{\mathrm{6}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:=\:\int\:\frac{{x}^{\mathrm{6}} \:+{x}^{\mathrm{4}} \:−{x}^{\mathrm{4}} \:−\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\:\int\:{x}^{\mathrm{4}} {dx}\:−\int\:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\:\int{x}^{\mathrm{4}} {dx}\:−\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:−\int\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{{x}^{\mathrm{5}} }{\mathrm{5}}\:−{arctanx}\:−\int\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{the}\:{ch}.{x}={tan}\theta\:{give} \\ $$$$\int\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\:\int\:\:\:\frac{{tan}^{\mathrm{4}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta=\:\int\:{tan}^{\mathrm{4}} \theta\:{d}\theta\:{but} \\ $$$$=\:\int\frac{\left.\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right)^{\mathrm{2}} }{d}\theta\:=\int\:\frac{\mathrm{1}−\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\:+\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}}{\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}\theta\right)+\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}}{d}\theta \\ $$$$=\int\:\:\:\:\:\frac{\mathrm{2}−\mathrm{4}{cos}\left(\mathrm{2}\theta\right)\:+\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}\:+\mathrm{4}{cos}\left(\mathrm{2}\theta\right)\:+\mathrm{1}\:+{cos}\left(\mathrm{4}\theta\right)}{d}\theta \\ $$$$=\int\:\:\:\frac{\mathrm{3}\:−\mathrm{4}{cos}\left(\mathrm{2}\theta\right)\:+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{3}\:+\mathrm{4}{cos}\left(\mathrm{2}\theta\right)\:+{cos}\left(\mathrm{4}\theta\right)}{d}\theta\:\:{this}\:{integral}\:{is}\:{calculable}\: \\ $$$${if}\:{the}\:{lmits}\:{are}\:{given}… \\ $$

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