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Question Number 134301 by bramlexs22 last updated on 02/Mar/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x^2}{{(1+x^2)}^4}dx$$

Answered by EDWIN88 last updated on 02/Mar/21

$$\mathrm{replace}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}\mathrm{yields}$$$$\mathrm{\Omega }={\int }_{\mathrm{\infty }}^0\frac{{\mathrm{x}}^{-2}}{{(1+{\mathrm{x}}^{-2})}^4}.(-{\mathrm{x}}^{-2}\mathrm{dx})={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^4}{{({\mathrm{x}}^2+4)}^4}\mathrm{dx}$$$$2\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^4+{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^4}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{(1+{\mathrm{x}}^2)}^3}\mathrm{dx}$$$$\mathrm{\Omega }=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{(1+{\mathrm{x}}^2)}^4}\mathrm{dx}.\mathrm{Apply}\mathrm{integration}\mathrm{by}\mathrm{parts}$$$$\{\begin{array}{l}\mathrm{u}=\frac{1}{2}\mathrm{x}\Rightarrow \mathrm{du}=\frac{1}{2}\mathrm{dx}\\ \mathrm{dv}=\frac{\mathrm{x}}{{({\mathrm{x}}^2+1)}^3}\mathrm{dx}\Rightarrow \mathrm{v}=-\frac{\mathrm{x}}{4{(1+{\mathrm{x}}^2)}^2}\end{array}$$$$\mathrm{\Omega }=-\frac{1}{8}\frac{{\mathrm{x}}^2}{{(1+{\mathrm{x}}^2)}^2}{\mid }_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{8{(1+{\mathrm{x}}^2)}^2}$$$$\mathrm{\Omega }=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}.\mathrm{replace}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}$$$$\mathrm{\Omega }=\frac{1}{8}{\int }_{\mathrm{\infty }}^0\frac{(-{\mathrm{x}}^{-2})}{{(1+{\mathrm{x}}^{-2})}^2}\mathrm{dx}=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}$$$$\mathrm{adding}\Rightarrow 2\mathrm{\Omega }=\frac{1}{8}{\int }_0^{\mathrm{\infty }}\frac{1+{\mathrm{x}}^2}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}$$$${\mathrm{\Omega }=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}=\frac{1}{16}\arctan \left(\mathrm{x}\right)]}_0^{\mathrm{\infty }}=\frac{\pi }{32}$$$$$$

Answered by Dwaipayan Shikari last updated on 02/Mar/21

$${\int }_0^{\mathrm{\infty }}\frac{x^2}{{(1+x^2)}^4}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}}}{{(1+u)}^4}du{x}^2=u$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{3}{2}-1}}{{(1+u)}^{\frac{5}{2}+\frac{3}{2}}}du=\frac{1}{2}.\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)\mathrm{\Gamma }\left(\frac{5}{2}\right)}{\mathrm{\Gamma }\left(4\right)}=\frac{1}{12}.\frac{1}{2}.\frac{3}{2}.\frac{1}{2}\pi =\frac{\pi }{32}$$

Answered by Ñï= last updated on 02/Mar/21

$$x=tan\theta$$$$\mathrm{\Omega }={\int }_0^{\pi /2}\frac{{tan}^2\theta }{{sec}^8\theta }{sec}^2\theta d\theta$$$$={\int }_0^{\pi /2}{cos}^4\theta {sin}^2\theta d\theta$$$$=\frac{1}{2}B(\frac{5}{2},\frac{3}{2})$$$$=\frac{\mathrm{\Gamma }\left(\frac{5}{2}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{2\mathrm{\Gamma }\left(4\right)}$$$$=\frac{\frac{3}{2}\times \frac{1}{2}\times \frac{1}{2}\pi }{2\times 3\times 2}$$$$=\frac{\pi }{32}$$

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