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Question Number 124061 by Bird last updated on 30/Nov/20

$$find{\int }_0^{\mathrm{\infty }}\frac{xarctanx}{{(x^2+1)}^2}dx$$

Answered by mnjuly1970 last updated on 30/Nov/20

$$\mathrm{\Omega }={\left[\frac{-1}{2(x^2+1)}{tan}^{-1}\left(x\right)\right]}_0^{\mathrm{\infty }}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{{(x^2+1)}^2}dx$$$$\stackrel{(x^2=\xi )}{=}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\xi }^{\frac{-1}{2}}}{{(1+\xi )}^2}d\xi =\frac{1}{4}\beta (\frac{1}{2},\frac{3}{2})$$$$=\frac{1}{4}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(2\right)}=\frac{\pi }{8}✓$$

Answered by Dwaipayan Shikari last updated on 30/Nov/20

$${\int }_0^{\mathrm{\infty }}\frac{{xtan}^{-1}x}{{(x^2+1)}^2}dx{tan}^{-1}x=\theta \Rightarrow \frac{1}{1+x^2}=\frac{d\theta }{dx}$$$$={\int }_0^{\frac{\pi }{2}}\frac{\theta tan\theta }{{sec}^2\theta }d\theta$$$$=-\frac{\theta }{4}{\left[cos\left(2\theta \right)\right]}_0^{\frac{\pi }{2}}+\frac{1}{8}{\left[sin2\theta \right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{8}$$

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