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x-3x-1-x-2-1-3-dx-

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Question Number 93804 by i jagooll last updated on 15/May/20
∫ x (((3x−1)/(x+2)))^(1/(3 )) dx ?
$$\int\:{x}\:\sqrt[{\mathrm{3}\:\:}]{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{2}}}\:{dx}\:?\: \\ $$
Commented by i jagooll last updated on 15/May/20
set t^3 = ((3x−1)/(x+2)) ⇒ xt^3 +2t^3 =3x−1 x=((−1−2t^3 )/(t^3 −3))= ((−2t^3 −1)/(t^3 −1)) dx = ((9t^2 )/((t^3 −1)^2 )) dt ⇒ ∫ (((−2t^3 −1)/(t^3 −3))).t (((9t^2 )/((t^3 −1)^2 ))) dt ∫ ((−9t^3 (2t^3 +1))/((t^3 −1)^2 (t^3 −3))) dt stuck???
$$\mathrm{set}\:\mathrm{t}^{\mathrm{3}} \:=\:\frac{\mathrm{3x}−\mathrm{1}}{\mathrm{x}+\mathrm{2}}\:\Rightarrow\:\mathrm{xt}^{\mathrm{3}} +\mathrm{2t}^{\mathrm{3}} =\mathrm{3x}−\mathrm{1}\: \\ $$$$\mathrm{x}=\frac{−\mathrm{1}−\mathrm{2t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{3}} −\mathrm{3}}=\:\frac{−\mathrm{2t}^{\mathrm{3}} −\mathrm{1}}{\mathrm{t}^{\mathrm{3}} −\mathrm{1}} \\ $$$$\mathrm{dx}\:=\:\frac{\mathrm{9t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dt}\: \\ $$$$\Rightarrow\:\int\:\left(\frac{−\mathrm{2t}^{\mathrm{3}} −\mathrm{1}}{\mathrm{t}^{\mathrm{3}} −\mathrm{3}}\right).\mathrm{t}\:\left(\frac{\mathrm{9t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} }\right)\:\mathrm{dt} \\ $$$$\int\:\frac{−\mathrm{9t}^{\mathrm{3}} \left(\mathrm{2t}^{\mathrm{3}} +\mathrm{1}\right)}{\left(\mathrm{t}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{3}} −\mathrm{3}\right)}\:\mathrm{dt}\: \\ $$$$\mathrm{stuck}??? \\ $$
Commented by mathmax by abdo last updated on 15/May/20
this integral is solved by a lots of calculus I =∫ x(^3 (√((3x−1)/(x+2))))dx changement^3 (√((3x−1)/(x+2)))=t give ((3x−1)/(x+2)) =t^3 ⇒3x−1 =t^3 x+2t^3 ⇒(3−t^3 )x =2t^3 +1 ⇒x =((2t^3 +1)/(3−t^3 )) ⇒(dx/dt) =((6t^2 (3−t^3 )−(2t^3 +1)(−3t^2 ))/((3−t^3 )^2 )) =((18t^2 −6t^5 +6t^5 −3t^2 )/((3−t^3 )^2 )) =((15t^2 )/((3−t^3 )^2 )) ⇒ I =∫ (((2t^3 +1)/(3−t^3 )))t ((15t^2 )/((3−t^3 )^2 )) dt = ∫ ((30t^5 +15t^2 )/((3−t^3 )^3 ))dt =−15∫ ((2t^5 +t^2 )/((t^3 −3)^3 ))dt let α =^3 (√3) ⇒ ∫ ((2t^5 +t^2 )/((t^3 −α^3 )^3 ))dt =∫ ((2t^5 +t^2 )/((t−α)^3 (t^2 +αt +α^2 )^3 ))dt =∫ ((2t^5 +t^2 )/((((t−α)/(t^2 +αt +α^2 )))^3 (t^2 +αt +α^2 )^6 ))dt changement ((t−α)/(t^2 +αt +α^2 )) =u give t+α =ut^2 +uαt +uα^2 ⇒ ut^2 +uαt −t−α+uα^2 =0 ⇒ ut^2 +(αu−1)t +uα^2 −α =0 Δ =(αu−1)^2 −4u(uα^2 −α) =α^2 u^2 −2αu −4α^2 u^2 +4αu =−3α^2 u^2 +2αu ⇒ t =((1−αu+^− (√(2αu−3α^2 u^2 )))/(2u)) ⇒ t =(1/(2u))−(α/2) +^− ((√(2αu−3α^2 u^2 ))/(2u)) ⇒ (dt/du) =−(1/(2u^2 )) +^− (1/2)((√(((2αu)/(4u^2 ))−((3α^2 u^2 )/(4u^2 )))))^′ =−(1/(2u^2 )) +^− (1/2)((√((α/(2u))−((3α^2 )/4))))^′ =−(1/(2u^2 )) +^− (1/2)(((−(α/(2u^2 )))/(2(√((α/(2u))−((3α^2 )/4)))))) =−(1/(2u^2 )) +^− (α/(8u^2 (√((α/(2u))−((3α^2 )/4))))) ....be continued....
$${this}\:{integral}\:{is}\:{solved}\:{by}\:{a}\:{lots}\:{of}\:{calculus}\: \\ $$$${I}\:=\int\:{x}\left(^{\mathrm{3}} \sqrt{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{2}}}\right){dx}\:{changement}\:^{\mathrm{3}} \sqrt{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{2}}}={t}\:{give} \\ $$$$\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{2}}\:={t}^{\mathrm{3}} \:\Rightarrow\mathrm{3}{x}−\mathrm{1}\:={t}^{\mathrm{3}} {x}+\mathrm{2}{t}^{\mathrm{3}} \:\Rightarrow\left(\mathrm{3}−{t}^{\mathrm{3}} \right){x}\:=\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{1}\:\Rightarrow{x}\:=\frac{\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{1}}{\mathrm{3}−{t}^{\mathrm{3}} } \\ $$$$\Rightarrow\frac{{dx}}{{dt}}\:=\frac{\mathrm{6}{t}^{\mathrm{2}} \left(\mathrm{3}−{t}^{\mathrm{3}} \right)−\left(\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{1}\right)\left(−\mathrm{3}{t}^{\mathrm{2}} \right)}{\left(\mathrm{3}−{t}^{\mathrm{3}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{18}{t}^{\mathrm{2}} −\mathrm{6}{t}^{\mathrm{5}} \:+\mathrm{6}{t}^{\mathrm{5}} −\mathrm{3}{t}^{\mathrm{2}} }{\left(\mathrm{3}−{t}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{15}{t}^{\mathrm{2}} }{\left(\mathrm{3}−{t}^{\mathrm{3}} \right)^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\int\:\left(\frac{\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{1}}{\mathrm{3}−{t}^{\mathrm{3}} }\right){t}\:\frac{\mathrm{15}{t}^{\mathrm{2}} }{\left(\mathrm{3}−{t}^{\mathrm{3}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$=\:\int\:\frac{\mathrm{30}{t}^{\mathrm{5}} \:+\mathrm{15}{t}^{\mathrm{2}} }{\left(\mathrm{3}−{t}^{\mathrm{3}} \right)^{\mathrm{3}} }{dt}\:=−\mathrm{15}\int\:\:\frac{\mathrm{2}{t}^{\mathrm{5}} \:+{t}^{\mathrm{2}} }{\left({t}^{\mathrm{3}} −\mathrm{3}\right)^{\mathrm{3}} }{dt}\:\:\:\:\:{let}\:\alpha\:=^{\mathrm{3}} \sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\int\:\:\frac{\mathrm{2}{t}^{\mathrm{5}} \:+{t}^{\mathrm{2}} }{\left({t}^{\mathrm{3}} −\alpha^{\mathrm{3}} \right)^{\mathrm{3}} }{dt}\:=\int\:\:\frac{\mathrm{2}{t}^{\mathrm{5}} \:+{t}^{\mathrm{2}} }{\left({t}−\alpha\right)^{\mathrm{3}} \left({t}^{\mathrm{2}} \:+\alpha{t}\:+\alpha^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$$$=\int\:\:\frac{\mathrm{2}{t}^{\mathrm{5}} \:+{t}^{\mathrm{2}} }{\left(\frac{{t}−\alpha}{{t}^{\mathrm{2}} \:+\alpha{t}\:+\alpha^{\mathrm{2}} }\right)^{\mathrm{3}} \:\left({t}^{\mathrm{2}} \:+\alpha{t}\:+\alpha^{\mathrm{2}} \right)^{\mathrm{6}} }{dt}\:\:{changement}\:\frac{{t}−\alpha}{{t}^{\mathrm{2}} \:+\alpha{t}\:+\alpha^{\mathrm{2}} }\:={u}\:{give} \\ $$$${t}+\alpha\:={ut}^{\mathrm{2}} \:+{u}\alpha{t}\:+{u}\alpha^{\mathrm{2}} \:\Rightarrow\:{ut}^{\mathrm{2}} \:+{u}\alpha{t}\:−{t}−\alpha+{u}\alpha^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow \\ $$$${ut}^{\mathrm{2}} \:+\left(\alpha{u}−\mathrm{1}\right){t}\:+{u}\alpha^{\mathrm{2}} −\alpha\:=\mathrm{0} \\ $$$$\Delta\:=\left(\alpha{u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{u}\left({u}\alpha^{\mathrm{2}} −\alpha\right)\:=\alpha^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{2}\alpha{u}\:−\mathrm{4}\alpha^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{4}\alpha{u} \\ $$$$=−\mathrm{3}\alpha^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{2}\alpha{u}\:\Rightarrow\:{t}\:=\frac{\mathrm{1}−\alpha{u}\overset{−} {+}\sqrt{\mathrm{2}\alpha{u}−\mathrm{3}\alpha^{\mathrm{2}} {u}^{\mathrm{2}} }}{\mathrm{2}{u}}\:\Rightarrow \\ $$$${t}\:=\frac{\mathrm{1}}{\mathrm{2}{u}}−\frac{\alpha}{\mathrm{2}}\:\overset{−} {+}\frac{\sqrt{\mathrm{2}\alpha{u}−\mathrm{3}\alpha^{\mathrm{2}} {u}^{\mathrm{2}} }}{\mathrm{2}{u}}\:\Rightarrow \\ $$$$\frac{{dt}}{{du}}\:=−\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }\:\overset{−} {+}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{2}\alpha{u}}{\mathrm{4}{u}^{\mathrm{2}} }−\frac{\mathrm{3}\alpha^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}{u}^{\mathrm{2}} }}\right)^{'} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }\:\overset{−} {+}\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\alpha}{\mathrm{2}{u}}−\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}}\right)^{'} \:=−\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }\:\overset{−} {+}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\frac{\alpha}{\mathrm{2}{u}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\frac{\alpha}{\mathrm{2}{u}}−\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }\:\overset{−} {+}\frac{\alpha}{\mathrm{8}{u}^{\mathrm{2}} \sqrt{\frac{\alpha}{\mathrm{2}{u}}−\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}}}\:\:\:….{be}\:{continued}…. \\ $$$$ \\ $$
Commented by i jagooll last updated on 15/May/20
very hard sir?
$$\mathrm{very}\:\mathrm{hard}\:\mathrm{sir}? \\ $$
Answered by MJS last updated on 15/May/20
∫x(((3x−1)/(x+2)))^(1/3) dx= [t=(((3x−1)/(x+2)))^(1/3) ⇔ x=−((2t^3 +1)/(t^3 −3)) → dx=((21t^2 )/((t^3 −3)^2 ))dt] =−21∫((t^3 (2t^3 +1))/((t^3 −3)^3 ))dt= [Ostrogradski] =((7t(43t^3 −66))/(18(t^3 −3)^2 ))−((77)/9)∫(dt/(t^3 −3)) now use this formula: ∫(dt/(t^3 −a^3 ))=(1/(3a^2 ))∫(dt/(t−a))−(1/(3a^2 ))∫((t+2a)/(t^2 +at+a^2 ))dt these should be easy to solve
$$\int{x}\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{2}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}+\mathrm{2}}}\:\Leftrightarrow\:{x}=−\frac{\mathrm{2}{t}^{\mathrm{3}} +\mathrm{1}}{{t}^{\mathrm{3}} −\mathrm{3}}\:\rightarrow\:{dx}=\frac{\mathrm{21}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{3}} −\mathrm{3}\right)^{\mathrm{2}} }{dt}\right] \\ $$$$=−\mathrm{21}\int\frac{{t}^{\mathrm{3}} \left(\mathrm{2}{t}^{\mathrm{3}} +\mathrm{1}\right)}{\left({t}^{\mathrm{3}} −\mathrm{3}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{\mathrm{7}{t}\left(\mathrm{43}{t}^{\mathrm{3}} −\mathrm{66}\right)}{\mathrm{18}\left({t}^{\mathrm{3}} −\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{77}}{\mathrm{9}}\int\frac{{dt}}{{t}^{\mathrm{3}} −\mathrm{3}} \\ $$$$\mathrm{now}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula}: \\ $$$$\int\frac{{dt}}{{t}^{\mathrm{3}} −{a}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\int\frac{{dt}}{{t}−{a}}−\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\int\frac{{t}+\mathrm{2}{a}}{{t}^{\mathrm{2}} +{at}+{a}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{these}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$
Commented by MJS last updated on 15/May/20
(1/(3a^2 ))∫(dt/(t−a))=(1/(3a^2 ))ln ∣t−a∣ (1/(3a^2 ))∫((t+2a)/(t^2 +at+a^2 ))dt=−(1/(6a^2 ))ln ∣t^2 +at+a^2 ∣ −((√3)/(3a^2 ))arctan (((√3)(2t+a))/(3a))
$$\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\int\frac{{dt}}{{t}−{a}}=\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\mathrm{ln}\:\mid{t}−{a}\mid \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} }\int\frac{{t}+\mathrm{2}{a}}{{t}^{\mathrm{2}} +{at}+{a}^{\mathrm{2}} }{dt}=−\frac{\mathrm{1}}{\mathrm{6}{a}^{\mathrm{2}} }\mathrm{ln}\:\mid{t}^{\mathrm{2}} +{at}+{a}^{\mathrm{2}} \mid\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}{a}^{\mathrm{2}} }\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+{a}\right)}{\mathrm{3}{a}} \\ $$

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