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Question Number 172973 by mnjuly1970 last updated on 04/Jul/22

$$$$$${\int }_0^{\mathrm{\infty }}\frac{1}{(1+x^4)(1+x^6)}dx=\frac{p\sqrt{2}-q}{12}\pi$$$$p,q=?$$$$$$

Answered by floor(10²Eta[1]) last updated on 04/Jul/22

$$$$$${\mathrm{x}}^6+1={\left({\mathrm{x}}^2\right)}^3+1=({\mathrm{x}}^2+1)({\mathrm{x}}^4-{\mathrm{x}}^2+1)$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^4+1)({\mathrm{x}}^2+1)({\mathrm{x}}^4-{\mathrm{x}}^2+1)}$$$$={\int }_0^{\mathrm{\infty }}(\frac{{\mathrm{x}}^2+1}{2({\mathrm{x}}^4+1)}+\frac{1}{6({\mathrm{x}}^2+1)}-\frac{{2\mathrm{x}}^2-1}{3({\mathrm{x}}^4-{\mathrm{x}}^2+1)})\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^4+1}\mathrm{dx}+\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}-\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{2\mathrm{x}}^2-1}{{\mathrm{x}}^4-{\mathrm{x}}^2+1}\mathrm{dx}$$$${◼=\mathrm{arctg}\left(\mathrm{x}\right)]}_0^{\mathrm{\infty }}=\frac{\pi }{2}$$$$◼={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+1}{{({\mathrm{x}}^2+1)}^2-{2\mathrm{x}}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2+1}{({\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1)({\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1)}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}(\frac{1}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1}+\frac{1}{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1})\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}+\frac{\sqrt{2}}{2})}^2+\frac{1}{2}}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}-\frac{\sqrt{2}}{2})}^2+\frac{1}{2}}$$$$=\frac{\sqrt{2}}{2}{\left[\mathrm{arctg}(\mathrm{x}\sqrt{2}+1)\right]}_0^{\mathrm{\infty }}+\frac{\sqrt{2}}{2}{\left[\mathrm{arctg}(\mathrm{x}\sqrt{2}-1)\right]}_0^{\mathrm{\infty }}$$$$=\frac{\sqrt{2}\pi }{2}$$$$◼={\mathrm{x}}^4-{\mathrm{x}}^2+1={({\mathrm{x}}^2+1)}^2-{3\mathrm{x}}^2$$$${\int }_0^{\mathrm{\infty }}\frac{{2\mathrm{x}}^2-1}{({\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1)({\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1)}\mathrm{dx}$$$$=\frac{1}{2\sqrt{3}}{\int }_0^{\mathrm{\infty }}(\frac{-3\mathrm{x}-\sqrt{3}}{{\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1}+\frac{3\mathrm{x}-\sqrt{3}}{{\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1})\mathrm{dx}$$$$=-\frac{\sqrt{3}}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+\sqrt{3}\mathrm{x}+1}+\frac{\sqrt{3}}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2-\sqrt{3}\mathrm{x}+1}$$$$=-\frac{\sqrt{3}}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{(\mathrm{x}+\frac{\sqrt{3}}{2})}^2+\frac{1}{4}}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}+\frac{\sqrt{3}}{2})}^2+\frac{1}{4}}+\frac{\sqrt{3}}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{(\mathrm{x}-\frac{\sqrt{3}}{2})}^2+\frac{1}{4}}\mathrm{dx}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}-\frac{\sqrt{3}}{2})}^2+\frac{1}{4}}$$$$=-2\sqrt{3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{(2\mathrm{x}+\sqrt{3})}^2+1}\mathrm{dx}-2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(2\mathrm{x}+\sqrt{3})}^2+1}+2\sqrt{3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{{(2\mathrm{x}-\sqrt{3})}^2+1}\mathrm{dx}-2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(2\mathrm{x}-\sqrt{3})}^2+1}$$$$\mathrm{u}=2\mathrm{x}\pm \sqrt{3}\Rightarrow \mathrm{du}=2\mathrm{dx},\mathrm{x}=\frac{\mathrm{u}\mp \sqrt{3}}{2}$$$$=-\frac{\sqrt{3}}{2}{\int }_{\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{u}-\sqrt{3}}{{\mathrm{u}}^2+1}\mathrm{du}-{\int }_{\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+1}+\frac{\sqrt{3}}{2}{\int }_{-\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{u}+\sqrt{3}}{{\mathrm{u}}^2+1}\mathrm{du}-{\int }_{-\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+1}$$$$-\frac{\sqrt{3}}{2}{\int }_{\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{u}}{{\mathrm{u}}^2+1}\mathrm{du}+\frac{1}{2}{\int }_{\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+1}+\frac{\sqrt{3}}{2}{\int }_{-\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{u}}{{\mathrm{u}}^2+1}\mathrm{du}+\frac{1}{2}{\int }_{-\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+1}$$$$-\frac{\sqrt{3}}{2}{\int }_{\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{u}}{{\mathrm{u}}^2+1}\mathrm{du}+\frac{\sqrt{3}}{2}{\int }_{-\sqrt{3}}^{\mathrm{\infty }}\frac{\mathrm{u}}{{\mathrm{u}}^2+1}\mathrm{du}+\frac{\pi }{2}$$$$\mathrm{t}={\mathrm{u}}^2+1\Rightarrow \mathrm{dt}=2\mathrm{udu}$$$$\frac{-\sqrt{3}}{4}{\int }_4^{\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{t}}+\frac{\sqrt{3}}{4}{\int }_4^{\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{t}}+\frac{\pi }{2}=\frac{\pi }{2}$$$$$$$$\Rightarrow \mathrm{I}=\frac{1}{2}◼+\frac{1}{6}◼-\frac{1}{3}◼$$$$=\frac{1}{2}\left(\frac{\pi \sqrt{2}}{2}\right)+\frac{1}{6}\left(\frac{\pi }{2}\right)-\frac{1}{3}\left(\frac{\pi }{2}\right)$$$$=\frac{\pi \sqrt{2}}{4}-\frac{\pi }{12}=\frac{3\sqrt{2}-1}{12}\pi \Rightarrow \mathrm{p}=3,\mathrm{q}=1$$

Commented by mnjuly1970 last updated on 04/Jul/22

$$\mathrm{bravo}\mathrm{sir}\mathrm{bravo}..$$

Commented by Tawa11 last updated on 06/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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