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Question Number 139328 by Bekzod Jumayev last updated on 25/Apr/21

Commented by Bekzod Jumayev last updated on 25/Apr/21

$$\mathit{H}\mathit{e}\mathit{l}\mathit{p}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}?$$$$$$

Answered by Ar Brandon last updated on 26/Apr/21

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^3+\mathrm{x}}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{\sqrt{\mathrm{x}}\cdot \sqrt{{\mathrm{x}}^2+1}},\mathrm{x}=\tan \theta \Rightarrow \mathrm{dx}={\sec }^2\theta \mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\theta \mathrm{d}\theta }{\sqrt{\tan \theta }\cdot \sqrt{{\sec }^2\theta }}={\int }_0^{\frac{\pi }{2}}\frac{\sec \theta }{\sqrt{\tan \theta }}\mathrm{d}\theta$$$$={\int }_0^{\frac{\pi }{2}}{\sin }^{-\frac{1}{2}}\theta {\cos }^{-\frac{1}{2}}\theta \mathrm{d}\theta =\frac{1}{2}\beta (\frac{1}{4},\frac{1}{4})=\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{2\mathrm{\Gamma }\left(\frac{1}{2}\right)}$$$$=\frac{1}{2\sqrt{\pi }}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$

Answered by Dwaipayan Shikari last updated on 26/Apr/21

$${\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{x^3+x}}{x}^2=u$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{4}-1}}{{(1+u)}^{\frac{1}{4}+\frac{1}{4}}}du=\frac{1}{2}\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}=\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{2\sqrt{\pi }}$$

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