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




Question Number 205294 by depressiveshrek last updated on 15/Mar/24
∫((3x−x^3 ))^(1/3) dx
$$\int\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−{x}^{\mathrm{3}} }{dx} \\ $$
Answered by Frix last updated on 15/Mar/24
∫(3x−x^3 )^(1/3) dx =^(t=(x^2 /3))  (3/2)∫t^(−(1/3)) (1−t)^(1/3) dt=  =(3/2)t^(2/3)  _2 F_1  (−(1/3), (2/3); (5/3); t) =  =(((3x^4 ))^(1/3) /2) _2 F_1  (−(1/3), (2/3); (5/3); (x^2 /3)) +C
$$\int\left(\mathrm{3}{x}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}\:\overset{{t}=\frac{{x}^{\mathrm{2}} }{\mathrm{3}}} {=}\:\frac{\mathrm{3}}{\mathrm{2}}\int{t}^{−\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dt}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{t}^{\frac{\mathrm{2}}{\mathrm{3}}} \:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{3}};\:\frac{\mathrm{5}}{\mathrm{3}};\:{t}\right)\:= \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}{x}^{\mathrm{4}} }}{\mathrm{2}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{2}}{\mathrm{3}};\:\frac{\mathrm{5}}{\mathrm{3}};\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)\:+{C} \\ $$

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