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Question Number 145636 by mathmax by abdo last updated on 06/Jul/21

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

Answered by Dwaipayan Shikari last updated on 06/Jul/21

$${\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}\left(x\right)}{{(1+x^2)}^2}dxx=tan\theta$$$$={\int }_0^{\frac{\pi }{2}}\frac{\theta }{{sec}^4\theta }{sec}^2\theta d\theta$$$$={\int }_0^{\frac{\pi }{2}}\theta {cos}^2\theta d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\theta +\theta cos2\theta d\theta$$$$=\frac{{\pi }^2}{16}-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}sin2\theta d\theta =\frac{{\pi }^2}{16}-\frac{1}{4}$$

Answered by mathmax by abdo last updated on 07/Jul/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{arctanx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}\mathrm{cha}7\mathrm{gement}\mathrm{x}=\mathrm{tant}\mathrm{give}$$$$\mathrm{\Psi }={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{t}}{{(1+{\tan }^2\mathrm{t})}^2}(1+{\tan }^2\mathrm{t})\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{t}}{1+{\tan }^2\mathrm{t}}\mathrm{dt}$$$$={\int }_0^{\frac{\pi }{2}}{\mathrm{tcos}}^2\mathrm{t}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\mathrm{t}\left(\frac{1+\cos \left(2\mathrm{t}\right)}{2}\right)\mathrm{dt}=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\mathrm{tdt}+\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\mathrm{tcos}\left(2\mathrm{t}\right)\mathrm{dt}$$$${\int }_0^{\frac{\pi }{2}}\mathrm{tdt}={\left[\frac{{\mathrm{t}}^2}{2}\right]}_0^{\frac{\pi }{2}}=\frac{{\pi }^2}{8}$$$${\int }_0^{\frac{\pi }{2}}\mathrm{tcos}\left(2\mathrm{t}\right)\mathrm{dt}={\left[\frac{\mathrm{t}}{2}\sin \left(2\mathrm{t}\right)\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\frac{\sin \left(2\mathrm{t}\right)}{2}\mathrm{dt}$$$$={\left[\frac{1}{4}\cos \left(2\mathrm{t}\right)\right]}_0^{\frac{\pi }{2}}=\frac{1}{4}(-2)=-\frac{1}{2}\Rightarrow$$$$\mathrm{\Psi }=\frac{{\pi }^2}{16}-\frac{1}{4}$$

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