Question Number 62855 by mathmax by abdo last updated on 26/Jun/19
$${find}\:\int\:\:\left(\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{6}} }\right)^{\mathrm{2}} \:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+{x}^{\mathrm{6}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+{x}^{\mathrm{6}} \right)^{\mathrm{2}} }{dx}\:. \\ $$
Commented by mathmax by abdo last updated on 27/Jun/19
![1) let I =∫ (x^8 /((1+x^6 )^2 ))dx ⇒I =∫ (x^8 /((1+(x^3 )^2 )^2 ))dx changement x^3 =tanθ give x =(tanθ)^(1/(3 )) ⇒ I =∫ (((tanθ)^(8/3) )/((1+tan^2 θ)^2 )) (1/3)(1+tan^2 θ)(tanθ)^(−(2/3)) dθ =(1/3) ∫ ((tan^2 θ)/(1+tan^2 θ)) dθ =(1/3) ∫ ((sin^2 θ)/(cos^2 θ)) cos^2 θ dθ =(1/3) ∫ sin^2 θ dθ =(1/3) ∫((1−cos(2θ))/2)dθ =(θ/6) −(1/(12)) sin(2θ) +c but θ =arctan(x^3 ) sin(2θ) =2sinθ cosθ =2 sin(arctan(x^3 ))cos(arctan(x^3 )) =2 (x^3 /(√(1+x^6 ))) .(1/(√(1+x^6 ))) =((2x^3 )/(1+x^6 )) ⇒ I =(1/6) arctan(x^3 )−(1/6)(x^3 /(1+x^6 )) +c 2) ∫_0 ^1 (x^8 /((1+x^6 )^2 ))dx =(1/6)[arctan(x^3 )−(x^3 /(1+x^6 ))]_0 ^1 =(1/6){(π/4) −(1/2)} 3) ∫_0 ^∞ (x^8 /((1+x^6 )^2 ))dx =(1/6)[ arctan(x^3 )−(x^3 /(1+x^6 ))]_0 ^(+∞) =(1/6) (π/2) =(π/(12)) . ×](Q63002.png)
$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\int\:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+{x}^{\mathrm{6}} \right)^{\mathrm{2}} }{dx}\:\Rightarrow{I}\:=\int\:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:{changement}\:{x}^{\mathrm{3}} ={tan}\theta\:{give} \\ $$$${x}\:=\left({tan}\theta\right)^{\frac{\mathrm{1}}{\mathrm{3}\:}} \:\Rightarrow\:{I}\:=\int\:\:\:\frac{\left({tan}\theta\right)^{\frac{\mathrm{8}}{\mathrm{3}}} }{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\left({tan}\theta\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} {d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\:\:\:\frac{{ta}\overset{\mathrm{2}} {{n}}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:{d}\theta\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\:\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int\:{sin}^{\mathrm{2}} \theta\:{d}\theta\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\theta}{\mathrm{6}}\:−\frac{\mathrm{1}}{\mathrm{12}}\:{sin}\left(\mathrm{2}\theta\right)\:+{c}\:\:\:{but}\:\theta\:={arctan}\left({x}^{\mathrm{3}} \right) \\ $$$${sin}\left(\mathrm{2}\theta\right)\:=\mathrm{2}{sin}\theta\:{cos}\theta\:=\mathrm{2}\:{sin}\left({arctan}\left({x}^{\mathrm{3}} \right)\right){cos}\left({arctan}\left({x}^{\mathrm{3}} \right)\right) \\ $$$$=\mathrm{2}\:\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}+{x}^{\mathrm{6}} }}\:.\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{6}} }}\:=\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{6}} }\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{\mathrm{6}}\:{arctan}\left({x}^{\mathrm{3}} \right)−\frac{\mathrm{1}}{\mathrm{6}}\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{6}} }\:+{c} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+{x}^{\mathrm{6}} \right)^{\mathrm{2}} }{dx}\:=\frac{\mathrm{1}}{\mathrm{6}}\left[{arctan}\left({x}^{\mathrm{3}} \right)−\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{6}} }\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{6}}\left\{\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+{x}^{\mathrm{6}} \right)^{\mathrm{2}} }{dx}\:=\frac{\mathrm{1}}{\mathrm{6}}\left[\:{arctan}\left({x}^{\mathrm{3}} \right)−\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{6}} }\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{6}}\:\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\mathrm{12}}\:. \\ $$$$× \\ $$