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Question Number 153152 by puissant last updated on 05/Sep/21

∫ (√(x^3 +1)) dx

$$\int\:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx} \\ $$

Answered by peter frank last updated on 05/Sep/21

u=(√(x^3 +1)) u^2 =x^3 +1 x^3 =(√(u^2 −1)) 2udu=3x^2 dx dx=2udu/3x^2 dx=2udu/3^3 (√(u^2 −1)) dx=((2udu)/(3(u^2 −1)^(1/3) )) ∫u.((2udu)/(3(u^2 −1)^(1/3) ))du (2/3)∫((u^2 du)/((u^2 −1)^(1/3) )) .....

$$\mathrm{u}=\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\mathrm{u}^{\mathrm{2}} =\mathrm{x}^{\mathrm{3}} +\mathrm{1}\:\:\mathrm{x}^{\mathrm{3}} =\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{2udu}=\mathrm{3x}^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{dx}=\mathrm{2udu}/\mathrm{3x}^{\mathrm{2}} \\ $$$$\mathrm{dx}=\mathrm{2udu}/\mathrm{3}^{\mathrm{3}} \sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{dx}=\frac{\mathrm{2udu}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\int\mathrm{u}.\frac{\mathrm{2udu}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }\mathrm{du} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{u}^{\mathrm{2}} \mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$..... \\ $$

Commented by peter frank last updated on 05/Sep/21

I failed to keep going (2/3)∫((u^2 du)/((u^2 −1)^(1/3) ))

$$\mathrm{I}\:\mathrm{failed}\:\mathrm{to}\:\mathrm{keep}\:\mathrm{going} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{u}^{\mathrm{2}} \mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$

Answered by aleks041103 last updated on 05/Sep/21

This integral doesn′t have solution in elementary functions.

$${This}\:{integral}\:{doesn}'{t}\:{have}\:{solution} \\ $$$${in}\:{elementary}\:{functions}. \\ $$

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