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Question Number 109616 by mathmax by abdo last updated on 24/Aug/20

$$\mathrm{find}{\int }_0^1\sqrt{1+{\mathrm{x}}^4}\mathrm{dx}$$

Commented by mathdave last updated on 24/Aug/20

$$thisishypergeometricquestionorquestion$$

Answered by mathmax by abdo last updated on 25/Aug/20

$$\mathrm{A}={\int }_0^1\sqrt{1+{\mathrm{x}}^4}\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{A}={\left[\mathrm{x}\sqrt{1+{\mathrm{x}}^4}\right]}_0^1-{\int }_0^1\mathrm{x}.\frac{{4\mathrm{x}}^3}{2\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}$$$$=\sqrt{2}-2{\int }_0^1\frac{{\mathrm{x}}^4}{\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}=\sqrt{2}-2{\int }_0^1\frac{{\mathrm{x}}^4+1-1}{\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}$$$$=\sqrt{2}-2{\int }_0^1\sqrt{1+{\mathrm{x}}^4}\mathrm{dx}+2{\int }_0^1\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^4)}^{\frac{1}{2}}}\Rightarrow$$$$3\mathrm{A}=\sqrt{2}+2{\int }_0^1\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^4)}^{\frac{1}{2}}}\mathrm{changement}\mathrm{x}={\mathrm{t}}^{\frac{1}{4}}\mathrm{give}$$$${\int }_0^1\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^4)}^{\frac{1}{2}}}=\frac{1}{4}{\int }_0^1\frac{{\mathrm{t}}^{\frac{1}{4}-1}}{{(1+\mathrm{t})}^{\frac{1}{2}}}\mathrm{dt}\mathrm{we}\mathrm{know}\mathrm{B}(\mathrm{p},\mathrm{q})={\int }_0^1\frac{{\mathrm{t}}^{\mathrm{p}-1}+{\mathrm{t}}^{\mathrm{q}-1}}{{(1+\mathrm{t})}^{\mathrm{p}+\mathrm{q}}}\mathrm{dt}$$$$=\frac{1}{8}{\int }_0^1\frac{{\mathrm{t}}^{\frac{1}{4}-1}+{\mathrm{t}}^{\frac{1}{4}-1}}{{(1+\mathrm{t})}^{\frac{1}{2}}}(\mathrm{p}=\mathrm{q}=\frac{1}{4})=\frac{1}{8}\mathrm{B}(\frac{1}{4},\frac{1}{4})=\frac{1}{8}.\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}$$$$=\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{8\sqrt{\pi }}\Rightarrow 3\mathrm{A}=\sqrt{2}+2\times \frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{8\sqrt{\pi }}=\sqrt{2}+\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{4\sqrt{\pi }}\Rightarrow$$$$\mathrm{A}=\frac{\sqrt{2}}{3}+\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{12\sqrt{\pi }}$$

Commented by mathdave last updated on 25/Aug/20

$$howdidyougetthe\frac{1}{8}whenitwas\frac{1}{4}{\int }_0^1\frac{t^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt$$$$has\frac{1}{2}didntfollowbetaformularyoustated$$

Commented by Aziztisffola last updated on 25/Aug/20

$$\frac{1}{4}{\int }_0^1\frac{t^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt=\frac{1}{2}\times (\frac{1}{4}{\int }_0^1\frac{t^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt+\frac{1}{4}{\int }_0^1\frac{t^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt)$$$$=\frac{1}{8}({\int }_0^1\frac{t^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt+{\int }_0^1\frac{t^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt)$$$$=\frac{1}{8}{\int }_0^1\frac{t^{\frac{1}{4}-1}+{\mathrm{t}}^{\frac{1}{4}-1}}{{(1+t)}^{\frac{1}{2}}}dt$$

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