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Question Number 186437 by aba last updated on 04/Feb/23

$$$$$$$$$${\int }_0^1\frac{1}{\sqrt{x\sqrt{x^2\sqrt{x^3\sqrt{x^4+1}}}}}dx$$$$$$

Commented by Frix last updated on 04/Feb/23

$$=\underset{0}{\stackrel{1}{\int }}\frac{dx}{x^{\frac{11}{8}}{(x^4+1)}^{\frac{1}{16}}}\mathrm{which}\mathrm{does}\mathrm{not}\mathrm{converge}$$$$\mathrm{The}\mathrm{indefinite}\mathrm{integral}\mathrm{is}$$$$-\frac{8}{3x^{\frac{3}{8}}}{\times }_2{F}_1(-\frac{3}{32},\frac{1}{16};\frac{29}{32};-x^4)+C$$

Commented by aba last updated on 04/Feb/23

$$\mathrm{thank}\mathrm{you}$$$$$$

Answered by a.lgnaoui last updated on 04/Feb/23

$$\frac{1}{\sqrt{x^2\sqrt{x3\sqrt{x^4+1}}}}=x{\left(\frac{1}{y}\right)}^2$$$$\frac{1}{x^2\sqrt{x^3\sqrt{x^4+1}}}={\left(\frac{x}{y^2}\right)}^2$$$$\frac{1}{\sqrt{x^3\sqrt{x^4+1}}}={\left(\frac{x}{y}\right)}^4$$$$\frac{1}{x^3\sqrt{x^4+1}}={\left(\frac{x}{y}\right)}^6$$$$\frac{1}{\sqrt{x^4+1}}=\frac{x^9}{y^6}\Rightarrow {\left(\frac{1}{y}\right)}^6=\frac{\sqrt{x^4+1}}{x^9}\left(1\right)$$$$\frac{2x^3}{\sqrt{x^4+1}}=\frac{d\left(\sqrt{x^4+1}\right)}{dx}\left(2\right)$$$$\frac{1}{y}=\frac{{(x^4+1)}^{-3}}{x^3}=\frac{1}{x^3{(x^4+1)}^3}$$$$$$$$\frac{1}{\sqrt{x\sqrt{x^2\sqrt{x^3\sqrt{x^4+1}}}}}=\frac{1}{x^3{(x^4+1)}^3}$$$$=\frac{2x^3}{{(x^4+1)}^3}\times \frac{1}{2x^6}=\frac{1}{2x^6}\frac{d\left(\sqrt{x^4+1}\right)}{dx}$$$$$$$$x^2=tdx=\frac{dt}{2x}=\frac{dt}{2\sqrt{t}}$$$$\frac{dt}{2t\sqrt{t}{(1+t^2)}^3}=\frac{dt}{2\left[\sqrt{t}{(1+t^2)}^3\right]}$$$$=\frac{1}{2{\left(\sqrt{t}\right)}^3}\times \frac{1}{{(1+t^2)}^3}$$$$U^{{}^{\prime }}=\frac{1}{2}{t}^{\frac{-3}{2}}V={\left(\frac{1}{1+t^2}\right)}^3$$$$U=-t^{-\frac{1}{2}}{V}^{{}^{\prime }}=-2t{(1+t^2)}^{-6}$$$$I=\frac{-\sqrt{t}}{{(1+t^2)}^3}+\int \frac{2t}{\sqrt{t}{(1+t^2)}^6}dt$$$$\frac{2t}{\sqrt{t}(1+t^2)}=-\int \frac{1}{\sqrt{t}}\times \frac{d}{dt}{\left(\frac{1}{1+t^2}\right)}^3$$$$I=\left[\frac{-\sqrt{t}}{1+t^2}\right]-2I$$$$3I={\left[\frac{-\sqrt{t}}{1+t^2}\right]}_0^1\Rightarrow I=\frac{1}{3}{\left[\frac{-\sqrt{t}}{1+t^2}\right]}_0^1$$$${\int }_0^1\frac{1}{\sqrt{x\sqrt{x^2\sqrt{x^3\sqrt{x^4+1}}}}}=-\frac{1}{3}$$

Commented by aba last updated on 04/Feb/23

$$\mathrm{thank}\mathrm{sir}$$$$\mathrm{error}\mathrm{line}5:\frac{1}{\sqrt{{\mathrm{x}}^4+1}}=\frac{{\mathrm{x}}^9}{{\mathrm{y}}^6}\Rightarrow {\left(\frac{1}{\mathrm{y}}\right)}^6=\frac{1}{{\mathrm{x}}^9\sqrt{{\mathrm{x}}^4+1}}=\frac{\sqrt{{\mathrm{x}}^4+1}}{{\mathrm{x}}^9({\mathrm{x}}^4+1)}$$

Commented by aba last updated on 05/Feb/23

$$\mathrm{the}\mathrm{integrale}\mathrm{does}\mathrm{not}\mathrm{converge}$$

Commented by Frix last updated on 05/Feb/23

$$\mathrm{Nonsense}\mathrm{again}.$$$$\sqrt{x\sqrt{x^2\sqrt{x^3\sqrt{x^4+1}}}}=x^{\frac{1}{2}}\sqrt[4]{x^2\sqrt{x^3\sqrt{x^4+1}}}=$$$$=x^{\frac{1}{2}}{x}^{\frac{1}{2}}\sqrt[8]{x^3\sqrt{x^4+1}}=x^{\frac{1}{2}}{x}^{\frac{1}{2}}{x}^{\frac{3}{8}}\sqrt[16]{x^4+1}=$$$$=x^{\frac{1}{2}+\frac{1}{2}+\frac{3}{8}}{(x^4+1)}^{\frac{1}{16}}=x^{\frac{11}{8}}{(x^4+1)}^{\frac{1}{16}}$$

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