dx-1-x-6-

Question Number 145324 by mathdanisur last updated on 04/Jul/21

$$\int\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{6}} }\:=\:? \\ $$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

(1/6)atan((x/(1−x^2 )))+(1/3)atanx+((√3)/(12))ln∣((x^2 +(√3)x+1)/(x^2 −(√3)x+1))∣+C

$$\frac{\mathrm{1}}{\mathrm{6}}\mathrm{atan}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{atan}{x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\mid\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}\mid+\mathrm{C} \\ $$

Commented by mathdanisur last updated on 04/Jul/21

$${Thanks}\:{Ser},\:{solution}\:{possible} \\ $$

Answered by Canebulok last updated on 04/Jul/21

Solution: let: ⇒ P^ = x^3 ⇛^3 (√P^( 2) ) = x^2 ⇒ dP = 3(^3 (√P^( 2) )) dx ∵ ⇒ (1/3)∙∫ (dP/((1+P^( 2) )(^3 (√P^( 2) )))) = ϕ By substitution, let: ⇒ u = P^( 2) ⇛ (√u) = P ⇒ du = 2P dP ⇛ (du/(2(√u))) = dP ∵ ⇒ (1/6)∙∫ (du/((1+u)(^6 (√u^5 )))) = ϕ By “I.B.P.”, let: ⇒ k = (1/(^6 (√u^5 ))) ⇒ dk = −(5/(6(^6 (√u^(11) )))) ⇒ dz = (du/((u+1))) ⇒ z = ln(u+1) ∵ ⇒ ϕ = ((ln(u+1))/(^6 (√u^5 ))) + ((5/6))∙∫ ((ln(u+1))/(^6 (√u^(11) ))) du ⇒ ϕ = 2tan^(−1) (^6 (√u))−2(^6 (√(−1)))^5 ∙tanh^(−1) (^6 (√(−u)))−2(^6 (√(−1)))∙tanh^(−1) ((−1)^(5/6) ∙(^6 (√u))))+C By substituting back again, ⇒ ϕ = 2tan^(−1) (^3 (√P))−2(^6 (√(−1)))^5 ∙tanh^(−1) (^6 (√(−1)) ∙^3 (√P) )−2(^6 (√(−1)))∙tanh^(−1) ((−1)^(5/6) ∙(^3 (√P)))+C Thus; ⇒ ϕ = 2tan^(−1) (x)−2(^6 (√(−1)))^5 ∙tanh^(−1) (^6 (√(−1)) ∙x )−2(^6 (√(−1)))∙tanh^(−1) ((−1)^(5/6) ∙(x))+C

$$\: \\ $$$$\boldsymbol{{Solution}}: \\ $$$${let}: \\ $$$$\Rightarrow\:{P}^{\:} \:=\:{x}^{\mathrm{3}} \:\Rrightarrow\:^{\mathrm{3}} \sqrt{{P}^{\:\mathrm{2}} }\:=\:{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{dP}\:=\:\mathrm{3}\left(^{\mathrm{3}} \sqrt{{P}^{\:\mathrm{2}} }\right)\:{dx} \\ $$$$\because \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{3}}\centerdot\int\:\frac{{dP}}{\left(\mathrm{1}+{P}^{\:\mathrm{2}} \right)\left(^{\mathrm{3}} \sqrt{{P}^{\:\mathrm{2}} }\right)}\:=\:\varphi \\ $$$$\: \\ $$$${By}\:{substitution}, \\ $$$${let}: \\ $$$$\Rightarrow\:{u}\:=\:{P}^{\:\mathrm{2}} \:\Rrightarrow\:\sqrt{{u}}\:=\:{P} \\ $$$$\Rightarrow\:{du}\:=\:\mathrm{2}{P}\:\:{dP}\:\Rrightarrow\:\frac{{du}}{\mathrm{2}\sqrt{{u}}}\:=\:{dP} \\ $$$$\because \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{6}}\centerdot\int\:\frac{{du}}{\left(\mathrm{1}+{u}\right)\left(^{\mathrm{6}} \sqrt{{u}^{\mathrm{5}} }\right)}\:=\:\varphi \\ $$$$\: \\ $$$${By}\:“{I}.{B}.{P}.'', \\ $$$${let}: \\ $$$$\Rightarrow\:{k}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{6}} \sqrt{{u}^{\mathrm{5}} }} \\ $$$$\Rightarrow\:{dk}\:=\:−\frac{\mathrm{5}}{\mathrm{6}\left(^{\mathrm{6}} \sqrt{{u}^{\mathrm{11}} }\right)} \\ $$$$\Rightarrow\:{dz}\:=\:\frac{{du}}{\left({u}+\mathrm{1}\right)} \\ $$$$\Rightarrow\:{z}\:=\:{ln}\left({u}+\mathrm{1}\right) \\ $$$$\because \\ $$$$\Rightarrow\:\varphi\:=\:\frac{{ln}\left({u}+\mathrm{1}\right)}{\:^{\mathrm{6}} \sqrt{{u}^{\mathrm{5}} }}\:+\:\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\centerdot\int\:\frac{{ln}\left({u}+\mathrm{1}\right)}{\:^{\mathrm{6}} \sqrt{{u}^{\mathrm{11}} }}\:{du} \\ $$$$\left.\Rightarrow\:\varphi\:=\:\:\mathrm{2tan}^{−\mathrm{1}} \left(^{\mathrm{6}} \sqrt{{u}}\right)−\mathrm{2}\left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\right)^{\mathrm{5}} \centerdot\mathrm{tanh}^{−\mathrm{1}} \left(^{\mathrm{6}} \sqrt{−{u}}\right)−\mathrm{2}\left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\right)\centerdot\mathrm{tanh}^{−\mathrm{1}} \left(\left(−\mathrm{1}\right)^{\mathrm{5}/\mathrm{6}} \centerdot\left(^{\mathrm{6}} \sqrt{{u}}\right)\right)\right)+{C} \\ $$$${By}\:{substituting}\:{back}\:{again}, \\ $$$$\Rightarrow\:\varphi\:=\:\mathrm{2tan}^{−\mathrm{1}} \left(^{\mathrm{3}} \sqrt{{P}}\right)−\mathrm{2}\left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\right)^{\mathrm{5}} \centerdot\mathrm{tanh}^{−\mathrm{1}} \left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\:\centerdot^{\mathrm{3}} \sqrt{{P}}\:\right)−\mathrm{2}\left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\right)\centerdot\mathrm{tanh}^{−\mathrm{1}} \left(\left(−\mathrm{1}\right)^{\mathrm{5}/\mathrm{6}} \centerdot\left(^{\mathrm{3}} \sqrt{{P}}\right)\right)+{C} \\ $$$${Thus}; \\ $$$$\Rightarrow\:\varphi\:=\:\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)−\mathrm{2}\left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\right)^{\mathrm{5}} \centerdot\mathrm{tanh}^{−\mathrm{1}} \left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\:\centerdot{x}\:\right)−\mathrm{2}\left(^{\mathrm{6}} \sqrt{−\mathrm{1}}\right)\centerdot\mathrm{tanh}^{−\mathrm{1}} \left(\left(−\mathrm{1}\right)^{\mathrm{5}/\mathrm{6}} \centerdot\left({x}\right)\right)+{C} \\ $$$$\: \\ $$

Commented by mathdanisur last updated on 04/Jul/21

$${Thanks}\:{Ser},\:{cool}\:{alot} \\ $$

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