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Question Number 201573 by sonukgindia last updated on 09/Dec/23

Answered by witcher3 last updated on 09/Dec/23

$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2024\mathrm{x}}+{\mathrm{e}}^{-2020}}{({\mathrm{e}}^{-2025\mathrm{x}}+{\mathrm{e}}^{-2019\mathrm{x}})({(-{4\mathrm{x}}^3+({4\mathrm{x}}^2+1)\sqrt{1+{\mathrm{x}}^2}-3\mathrm{x})}^{2023}+1)}=\mathrm{I}$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{5\mathrm{x}}}{(1+{\mathrm{e}}^{6\mathrm{x}}){\left((-{4\mathrm{x}}^3-3\mathrm{x}+({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1})\left(\frac{-{4\mathrm{x}}^3-3\mathrm{x}-({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1}}{-{4\mathrm{x}}^3-3\mathrm{x}-({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1}}\right)\right)}^2+1)}$$$$\{({4\mathrm{x}}^3+3\mathrm{x})+({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1}\}\{.({4\mathrm{x}}^3+3\mathrm{x})-({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1\}}=-1$$$$\mathrm{I}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{({\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{5\mathrm{x}}){({4\mathrm{x}}^3+3\mathrm{x}+({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1})}^{2023}}{(1+{\mathrm{e}}^{6\mathrm{x}})(1+{({4\mathrm{x}}^3+3\mathrm{x}+({4\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+1})}^{2023}}\mathrm{dx}$$$$2\mathrm{I}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{5\mathrm{x}}}{1+{\mathrm{e}}^{6\mathrm{x}}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{1+{\mathrm{x}}^4}{1+{\mathrm{x}}^6}\mathrm{dx},{\mathrm{x}}^6=\mathrm{y}$$$$={\int }_0^{\mathrm{\infty }}\frac{1+{\mathrm{y}}^{\frac{2}{3}}}{1+\mathrm{y}}.\frac{{\mathrm{y}}^{-\frac{5}{6}}}{6}\mathrm{dy}$$$$=\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{-\frac{5}{6}}}{1+\mathrm{y}}\mathrm{dy}+\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{y}}^{-\frac{1}{6}}}{1+\mathrm{y}}\mathrm{dy}$$$$=\frac{1}{6}\beta (\frac{1}{6},\frac{5}{6})+\frac{1}{6}\beta (\frac{5}{6},\frac{1}{6})=\frac{1}{3}.\frac{\pi }{\sin \left(\frac{\pi }{6}\right)}$$$$=\frac{2\pi }{3}$$$$\mathrm{I}=\frac{\pi }{3}$$$$$$

Commented by Anonim_X last updated on 09/Dec/23

nice

Commented by Calculusboy last updated on 09/Dec/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$

Commented by witcher3 last updated on 10/Dec/23

$$\mathrm{thank}\mathrm{You}$$

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