Menu Close

dx-x-4-x-2-a-2-

Tinku Tara 0 Comments
Pin




Question Number 138296 by bobhans last updated on 12/Apr/21
∫ (dx/(x^4 (√(x^2 −a^2 )))) =?
$$\int\:\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:=? \\ $$
Answered by bemath last updated on 12/Apr/21
Answered by mathmax by abdo last updated on 12/Apr/21
I=∫ (dx/(x^4 (√(x^2 −a^2 )))) we do the changement x=ach(t) ⇒ I=∫ ((ash(t))/(a^4 ch^4 t(asht)))dt =(1/a^4 ) ∫ (dt/((((1+ch(2t))/2))^2 )) =(4/a^4 )∫ (dt/(1+2ch(2t)+ch^2 (2t))) =(4/a^4 )∫ (dt/(1+2ch(2t)+((1+ch(4t))/2))) =(8/a^3 )∫ (dt/(2+4ch(2t)+1+ch(4t))) =_(2t=u) (8/a^3 )∫ (du/(2(3+4ch(u)+ch(2u)))) =(4/a^3 )∫ (du/(ch(2u)+4chu +3)) =(4/a^3 )∫ (du/(((e^(2u) +e^(−2u) )/2)+4((e^u +e^(−u) )/2)+3)) =(8/a^3 )∫ (du/(e^(2u) +e^(−2u) +4e^u +4e^(−u) +6)) =_(e^(u ) =y) (8/a^3 )∫ (dy/(y(y^2 +y^(−2) +4y+4y^(−1) +6))) =(8/a^3 )∫ (dy/(y^3 +y^(−1) +4y^2 +4 +6y))=(8/a^3 )∫ ((ydy)/(y^4 +1+4y^3 +4y +6y^2 )) rest to decompose F(y)=(y/(y^4 +4y^3 +6y^2 +4y +1)) ....be continued....
$$\mathrm{I}=\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{ach}\left(\mathrm{t}\right)\:\Rightarrow \\ $$$$\mathrm{I}=\int\:\:\frac{\mathrm{ash}\left(\mathrm{t}\right)}{\mathrm{a}^{\mathrm{4}} \mathrm{ch}^{\mathrm{4}} \mathrm{t}\left(\mathrm{asht}\right)}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{4}} }\:\int\:\:\frac{\mathrm{dt}}{\left(\frac{\mathrm{1}+\mathrm{ch}\left(\mathrm{2t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{4}} }\int\:\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{2ch}\left(\mathrm{2t}\right)+\mathrm{ch}^{\mathrm{2}} \left(\mathrm{2t}\right)}\:=\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{4}} }\int\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{2ch}\left(\mathrm{2t}\right)+\frac{\mathrm{1}+\mathrm{ch}\left(\mathrm{4t}\right)}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{8}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\frac{\mathrm{dt}}{\mathrm{2}+\mathrm{4ch}\left(\mathrm{2t}\right)+\mathrm{1}+\mathrm{ch}\left(\mathrm{4t}\right)}\:=_{\mathrm{2t}=\mathrm{u}} \:\:\frac{\mathrm{8}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\:\:\frac{\mathrm{du}}{\mathrm{2}\left(\mathrm{3}+\mathrm{4ch}\left(\mathrm{u}\right)+\mathrm{ch}\left(\mathrm{2u}\right)\right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\:\:\:\frac{\mathrm{du}}{\mathrm{ch}\left(\mathrm{2u}\right)+\mathrm{4chu}\:+\mathrm{3}}\:=\frac{\mathrm{4}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\frac{\mathrm{du}}{\frac{\mathrm{e}^{\mathrm{2u}} +\mathrm{e}^{−\mathrm{2u}} }{\mathrm{2}}+\mathrm{4}\frac{\mathrm{e}^{\mathrm{u}} +\mathrm{e}^{−\mathrm{u}} }{\mathrm{2}}+\mathrm{3}} \\ $$$$=\frac{\mathrm{8}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\:\frac{\mathrm{du}}{\mathrm{e}^{\mathrm{2u}} +\mathrm{e}^{−\mathrm{2u}} \:+\mathrm{4e}^{\mathrm{u}} \:+\mathrm{4e}^{−\mathrm{u}} \:+\mathrm{6}} \\ $$$$=_{\mathrm{e}^{\mathrm{u}\:} =\mathrm{y}} \:\:\:\frac{\mathrm{8}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\:\frac{\mathrm{dy}}{\mathrm{y}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{y}^{−\mathrm{2}} \:+\mathrm{4y}+\mathrm{4y}^{−\mathrm{1}} \:+\mathrm{6}\right)} \\ $$$$=\frac{\mathrm{8}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{3}} +\mathrm{y}^{−\mathrm{1}} \:+\mathrm{4y}^{\mathrm{2}} \:+\mathrm{4}\:+\mathrm{6y}}=\frac{\mathrm{8}}{\mathrm{a}^{\mathrm{3}} }\int\:\:\frac{\mathrm{ydy}}{\mathrm{y}^{\mathrm{4}} \:+\mathrm{1}+\mathrm{4y}^{\mathrm{3}} \:+\mathrm{4y}\:+\mathrm{6y}^{\mathrm{2}} } \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{4}} \:+\mathrm{4y}^{\mathrm{3}} \:+\mathrm{6y}^{\mathrm{2}} \:+\mathrm{4y}\:+\mathrm{1}} \\ $$$$….\mathrm{be}\:\mathrm{continued}…. \\ $$
Answered by Ñï= last updated on 12/Apr/21
I=∫(dx/(x^4 (√(x^2 −a^2 )))) x=asec θ I=∫((atan θsec θ)/(a^5 sec^4 θtan θ))dθ=(1/a^4 )∫cos^3 θdθ=(1/a^4 )∫(1−sin^2 θ)d(sin θ) =(1/a^4 )(sin θ−(1/3)sin^3 θ)+C =(1/a^4 )((((x^2 −a^2 )^(1/2) )/x)−(((x^2 −a^2 )^(3/2) )/(3x^3 )))+C
$${I}=\int\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$${x}={a}\mathrm{sec}\:\theta \\ $$$${I}=\int\frac{{a}\mathrm{tan}\:\theta\mathrm{sec}\:\theta}{{a}^{\mathrm{5}} \mathrm{sec}\:^{\mathrm{4}} \theta\mathrm{tan}\:\theta}{d}\theta=\frac{\mathrm{1}}{{a}^{\mathrm{4}} }\int\mathrm{cos}\:^{\mathrm{3}} \theta{d}\theta=\frac{\mathrm{1}}{{a}^{\mathrm{4}} }\int\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\right){d}\left(\mathrm{sin}\:\theta\right) \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{4}} }\left(\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:^{\mathrm{3}} \theta\right)+{C} \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{4}} }\left(\frac{\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} }{{x}}−\frac{\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}{x}^{\mathrm{3}} }\right)+{C} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *