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Question Number 142103 by rs4089 last updated on 26/May/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{tan}^{-1}\left(\sqrt{x^2+2}\right)}{(x^2+1)\sqrt{x^2+2}}dx$$

Answered by mathmax by abdo last updated on 26/May/21

$$\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan \left(\mathrm{a}\sqrt{{\mathrm{x}}^2+2}\right)}{({\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+2}}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+1)(1+{\mathrm{a}}^2({\mathrm{x}}^2+2))}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+1)({\mathrm{a}}^2{\mathrm{x}}^2+{2\mathrm{a}}^2+1)}$$$${=}_{\mathrm{ax}=\mathrm{t}}2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{a}(\frac{{\mathrm{t}}^2}{{\mathrm{a}}^2}+1)({\mathrm{t}}^2+{2\mathrm{a}}^2+1)}=2\mathrm{a}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{({\mathrm{t}}^2+{\mathrm{a}}^2)({\mathrm{t}}^2+{2\mathrm{a}}^2+1)}$$$$=\frac{2\mathrm{a}}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}(\frac{1}{{\mathrm{t}}^2+{\mathrm{a}}^2}-\frac{1}{{\mathrm{t}}^2+{2\mathrm{a}}^2+1})\mathrm{dt}$$$$=\frac{2\mathrm{a}}{{\mathrm{a}}^2+1}\{{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+{\mathrm{a}}^2}-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+{2\mathrm{a}}^2+1}\}\mathrm{we}\mathrm{have}$$$${{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+{\mathrm{a}}^2}{=}_{\mathrm{t}=\mathrm{ay}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{ady}}{{\mathrm{a}}^2({\mathrm{y}}^2+1)}=\frac{1}{\mathrm{a}}\arctan \left(\frac{\mathrm{t}}{\mathrm{a}}\right)]}_0^{\mathrm{\infty }}=\frac{\pi }{2\mathrm{a}}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+{2\mathrm{a}}^2+1}{=}_{\mathrm{t}=\sqrt{{2\mathrm{a}}^2+1}\mathrm{y}}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{{2\mathrm{a}}^2+1}}{({2\mathrm{a}}^2+1)({\mathrm{y}}^2+1)}\mathrm{dy}$$$${=\frac{1}{\sqrt{{2\mathrm{a}}^2+1}}\arctan \left(\frac{\mathrm{t}}{\sqrt{{2\mathrm{a}}^2+1}}\right)]}_0^{\mathrm{\infty }}=\frac{\pi }{2\sqrt{{2\mathrm{a}}^2+1}}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{2\mathrm{a}}{{\mathrm{a}}^2+1}\frac{\pi }{2\mathrm{a}}-\frac{2\mathrm{a}}{{\mathrm{a}}^2+1}\frac{\pi }{2\sqrt{{2\mathrm{a}}^2+1}}=\frac{\pi }{{\mathrm{a}}^2+1}-\frac{\mathrm{a}\pi }{({\mathrm{a}}^2+1)\sqrt{{2\mathrm{a}}^2+1}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\pi {\int }_0^{\mathrm{a}}\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}-\pi {\int }_0^{\mathrm{a}}\frac{\mathrm{xdx}}{({\mathrm{x}}^2+1)\sqrt{{2\mathrm{x}}^2+1}}+\mathrm{C}$$$$\mathrm{f}\left(0\right)=0=\mathrm{C}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\pi {\int }_0^{\mathrm{a}}\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}-\pi {\int }_0^{\mathrm{a}}\frac{\mathrm{xdx}}{({\mathrm{x}}^2+1)\sqrt{{2\mathrm{x}}^2+1}}$$$$\mathrm{and}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan \left(\sqrt{{\mathrm{x}}^2+2}\right)}{({\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^2+2}}\mathrm{dx}=\mathrm{f}\left(1\right)=\pi {\int }_0^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}-\pi {\int }_0^1\frac{\mathrm{xdx}}{({\mathrm{x}}^2+1)\sqrt{{2\mathrm{x}}^2+1}}$$$$\mathrm{we}\mathrm{have}{\int }_0^1\frac{\mathrm{dx}}{{\mathrm{x}}^2+1}=\frac{\pi }{4}$$$${\int }_0^1\frac{\mathrm{xdx}}{({\mathrm{x}}^2+1)\sqrt{{2\mathrm{x}}^2+1}}{=}_{\sqrt{2}\mathrm{x}=\mathrm{shy}}{\int }_0^{\mathrm{argsh}\left(\sqrt{2}\right)}\frac{\mathrm{shy}}{\sqrt{2}(\frac{{\mathrm{sh}}^2\mathrm{y}}{2}+1)\mathrm{chy}}\frac{\mathrm{chy}}{\sqrt{2}}\mathrm{dy}$$$$=\frac{1}{2}{\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{\mathrm{shy}}{\frac{1}{2}{\mathrm{sh}}^2\mathrm{y}+1}\mathrm{dy}={\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{\mathrm{shy}}{{\mathrm{sh}}^2\mathrm{y}+2}\mathrm{dy}$$$$={\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{\mathrm{shy}}{\frac{\mathrm{ch}\left(2\mathrm{y}\right)-1}{2}+2}\mathrm{dy}=2{\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{\mathrm{shy}}{\mathrm{ch}\left(2\mathrm{y}\right)+3}\mathrm{dy}$$$$=2{\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{\frac{{\mathrm{e}}^{\mathrm{y}}-{\mathrm{e}}^{-\mathrm{y}}}{2}}{\frac{{\mathrm{e}}^{2\mathrm{y}}-{\mathrm{e}}^{-2\mathrm{y}}}{2}+3}\mathrm{dy}=2{\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{{\mathrm{e}}^{\mathrm{y}}-{\mathrm{e}}^{-\mathrm{y}}}{{\mathrm{e}}^{2\mathrm{y}}-{\mathrm{e}}^{-2\mathrm{y}}+6}\mathrm{dy}$$$$=2{\int }_0^{\log (\sqrt{2}+\sqrt{3})}\frac{{\mathrm{e}}^{3\mathrm{y}}-{\mathrm{e}}^{\mathrm{y}}}{{\mathrm{e}}^{4\mathrm{y}}-1+{6\mathrm{e}}^{2\mathrm{y}}}\mathrm{dy}$$$${=}_{{\mathrm{e}}^{\mathrm{y}}=\mathrm{u}}2{\int }_1^{\sqrt{2}+\sqrt{3}}\frac{{\mathrm{u}}^3-\mathrm{u}}{{\mathrm{u}}^4+{6\mathrm{u}}^2-1}\frac{\mathrm{du}}{\mathrm{u}}$$$$=2{\int }_1^{\sqrt{2}+\sqrt{3}}\frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^4+{6\mathrm{u}}^2-1}\mathrm{du}$$$${\mathrm{u}}^4+{6\mathrm{u}}^2-1=0\Rightarrow {\mathrm{z}}^2+6\mathrm{z}-1=0(\mathrm{z}={\mathrm{u}}^2)$$$${\mathrm{\Delta }}^{{}^{\prime }}=9+1=10\Rightarrow {\mathrm{u}}_1=-3+\sqrt{10}\mathrm{and}{\mathrm{u}}_2=-3-\sqrt{10}$$$${\mathrm{u}}^4+{6\mathrm{u}}^2-1=({\mathrm{u}}^2-{\mathrm{u}}_1)({\mathrm{u}}^2-{\mathrm{u}}_2)\Rightarrow$$$$\mathrm{\Psi }\left(\mathrm{u}\right)=\frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^4+{6\mathrm{u}}^2-1}=({\mathrm{u}}^2-1)(\frac{1}{{\mathrm{u}}^2-{\mathrm{u}}_1}-\frac{1}{{\mathrm{u}}^2-{\mathrm{u}}_2})\times \frac{1}{{\mathrm{u}}_1-{\mathrm{u}}_2}$$$$=\frac{1}{2\sqrt{10}}(\frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^2-{\mathrm{u}}_1}-\frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^2-{\mathrm{u}}_2})=\frac{1}{2\sqrt{10}}(\frac{{\mathrm{u}}^2-{\mathrm{u}}_1+{\mathrm{u}}_1-1}{{\mathrm{u}}^2-{\mathrm{u}}_1}-\frac{{\mathrm{u}}^2-{\mathrm{u}}_2+{\mathrm{u}}_2-1}{{\mathrm{u}}^2-{\mathrm{u}}_2})$$$$=\frac{1}{2\sqrt{10}}(\frac{{\mathrm{u}}_1-1}{{\mathrm{u}}^2-{\mathrm{u}}_1}-\frac{{\mathrm{u}}_2-1}{{\mathrm{u}}^2-{\mathrm{u}}_2})\Rightarrow$$$${\int }_0^1\frac{\mathrm{xdx}}{({\mathrm{x}}^2+1)\sqrt{{2\mathrm{x}}^2+1}}=\frac{{\mathrm{u}}_1-1}{\sqrt{10}}{\int }_1^{\sqrt{2}+\sqrt{3}}\frac{\mathrm{du}}{{\mathrm{u}}^2-{\mathrm{u}}_1}-\frac{{\mathrm{u}}_2-1}{\sqrt{10}}{\int }_1^{\sqrt{2}+\sqrt{3}}\frac{\mathrm{du}}{{\mathrm{u}}^2-{\mathrm{u}}_2}$$$$\mathrm{now}\mathrm{its}\mathrm{eazy}\mathrm{to}\mathrm{find}\mathrm{those}\mathrm{integrals}....$$

Answered by mindispower last updated on 26/May/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{tan}^{-}\left(t\sqrt{x^2+2}\right)}{\sqrt{x^2+2}(1+x^2)}dx=f\left(t\right)$$$$f^{\prime }\left(t\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{(1+x^2)(1+2t^2+t^2{x}^2)}$$$$=2i\pi .(\frac{1}{2i(1+t^2)}+\frac{1}{1-\frac{1+2t^2}{t^2}}.\frac{1}{2it\sqrt{1+2t^2}})$$$$=\frac{\pi }{1+t^2}+\frac{\pi t}{-(1+t^2)\sqrt{1+2t^2}}=f^{\prime }\left(t\right)$$$$f\left(0\right)=0\iff {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{arctan\left(\sqrt{x^2+2}\right)}{(1+x^2)\sqrt{x^2+2}}dx=f\left(1\right)={\int }_0^1{f}^{\prime }\left(t\right)dt$$$$={\int }_0^1\frac{\pi }{1+t^2}-\frac{\pi }{2}{\int }_0^1\frac{dx}{(1+x)\sqrt{1+2x}}dx$$$$=\frac{{\pi }^2}{4}-\frac{\pi }{2}{\int }_1^3\frac{udu}{\frac{u^2+1}{2}u}=\frac{{\pi }^2}{4}-\pi {\int }_1^{\sqrt{3}}\frac{du}{(u^2+1)}$$$$=\frac{{\pi }^2}{4}-\pi \left[{}_1^{\sqrt{3}}arctan\left(u\right)\right]$$$$=\frac{{\pi }^2}{4}-\pi .\frac{\pi }{12}=\frac{{\pi }^2}{6}$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(\sqrt{2x+1}\right)}{(1+x^2)\sqrt{2x+1}}dx=\frac{{\pi }^2}{6}$$$$$$

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