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Question Number 84170 by jagoll last updated on 10/Mar/20

∫ (dx/(x^3 ((4−x^3 ))^(1/(3 )) )) ?

$$\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \:\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{4}−\mathrm{x}^{\mathrm{3}} }}\:?\: \\ $$

Answered by john santu last updated on 10/Mar/20

∫ (dx/(x^4 (((4/x^3 )−1))^(1/(3 )) )) = let u = (4/x^3 )−1 ⇒ du = −((12)/x^4 ) dx (dx/x^4 ) = −(1/(12))du. ⇒∫ −(du/(12u^(1/3) )) = −(1/(12))×(3/2)(u^2 )^(1/(3 )) + c =−(1/8)(((((4−x^3 )/x^3 ))^2 ))^(1/(3 )) + c −(1/(8x^2 )) (((4−x^3 )^2 ))^(1/(3 )) + c

$$\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} \:\sqrt[{\mathrm{3}\:\:}]{\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{3}} }−\mathrm{1}}}\:=\: \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{3}} }−\mathrm{1}\:\Rightarrow\:\mathrm{du}\:=\:−\frac{\mathrm{12}}{\mathrm{x}^{\mathrm{4}} }\:\mathrm{dx} \\ $$$$\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} }\:=\:−\frac{\mathrm{1}}{\mathrm{12}}\mathrm{du}. \\ $$$$\Rightarrow\int\:−\frac{{du}}{\mathrm{12}{u}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:\:=\:−\frac{\mathrm{1}}{\mathrm{12}}×\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}\:\:}]{{u}^{\mathrm{2}} }\:+\:{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\sqrt[{\mathrm{3}\:\:}]{\left(\frac{\mathrm{4}−{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} }\:+\:{c} \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }\:\sqrt[{\mathrm{3}\:\:}]{\left(\mathrm{4}−{x}^{\mathrm{3}} \right)^{\mathrm{2}} }\:+\:{c} \\ $$

Answered by MJS last updated on 10/Mar/20

∫(dx/(x^3 ((4−x^3 ))^(1/3) ))=−∫(dx/(x^3 ((x^3 −4))^(1/3) ))= [t=(((x^3 −4))^(1/3) /x) → dx=((x^2 (x^3 −4)^(2/3) )/4)] =−(1/4)∫tdt=−(1/8)t^2 =−(((x^3 −4)^(2/3) )/(8x^2 ))+C

$$\int\frac{{dx}}{{x}^{\mathrm{3}} \sqrt[{\mathrm{3}}]{\mathrm{4}−{x}^{\mathrm{3}} }}=−\int\frac{{dx}}{{x}^{\mathrm{3}} \sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} −\mathrm{4}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} −\mathrm{4}}}{{x}}\:\rightarrow\:{dx}=\frac{{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{4}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{4}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int{tdt}=−\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{2}} =−\frac{\left({x}^{\mathrm{3}} −\mathrm{4}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{8}{x}^{\mathrm{2}} }+{C} \\ $$

Commented by john santu last updated on 10/Mar/20

oo thank you sir

$$\mathrm{oo}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by john santu last updated on 10/Mar/20

how to get t = (((x^3 −4))^(1/(3 )) /x) sir? by feeling?

$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{t}\:=\:\frac{\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}^{\mathrm{3}} −\mathrm{4}}}{\mathrm{x}}\:\mathrm{sir}? \\ $$$$\mathrm{by}\:\mathrm{feeling}? \\ $$

Commented by MJS last updated on 10/Mar/20

I tried t=((x^3 −4))^(1/3) → dx=(((x^3 −4)^(2/3) )/(3x^2 )) first and because of ((u/v))^′ =((u′v−uv′)/v^2 ) with u=(w)^(1/3) is (((1/3)w^(−2/3) v−w^(1/3) v′)/v^2 )=((v−wv′)/(3v^2 w^(2/3) )) I tried v=x maybe it′s just experience mixed with good luck

$$\mathrm{I}\:\mathrm{tried}\:{t}=\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} −\mathrm{4}}\:\rightarrow\:{dx}=\frac{\left({x}^{\mathrm{3}} −\mathrm{4}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{2}} }\:\mathrm{first}\:\mathrm{and} \\ $$$$\mathrm{because}\:\mathrm{of}\:\left(\frac{{u}}{{v}}\right)^{'} =\frac{{u}'{v}−{uv}'}{{v}^{\mathrm{2}} }\:\mathrm{with}\:{u}=\sqrt[{\mathrm{3}}]{{w}}\:\mathrm{is} \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{3}}{w}^{−\mathrm{2}/\mathrm{3}} {v}−{w}^{\mathrm{1}/\mathrm{3}} {v}'}{{v}^{\mathrm{2}} }=\frac{{v}−{wv}'}{\mathrm{3}{v}^{\mathrm{2}} {w}^{\mathrm{2}/\mathrm{3}} }\:\mathrm{I}\:\mathrm{tried}\:{v}={x} \\ $$$$\mathrm{maybe}\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{experience}\:\mathrm{mixed}\:\mathrm{with}\:\mathrm{good}\:\mathrm{luck} \\ $$

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