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Question Number 148467 by bramlexs22 last updated on 28/Jul/21

Answered by puissant last updated on 28/Jul/21

$$\mathrm{x}=\tan \left(\mathrm{t}\right)\Rightarrow \mathrm{dx}=1+{\tan }^2\left(\mathrm{t}\right)\mathrm{dt}$$$$0⩽\mathrm{x}⩽\mathrm{\infty }\Rightarrow 0⩽\mathrm{t}⩽\frac{\pi }{2}$$$$\Rightarrow \mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\arctan \left(\tan \left(\mathrm{t}\right)\right)}{1+{\tan }^2\left(\mathrm{t}\right)}(1+{\tan }^2\left(\mathrm{t}\right))\mathrm{dt}$$$$={\int }_0^{\frac{\pi }{2}}\mathrm{tdt}=\frac{1}{2}{\left[{\mathrm{t}}^2\right]}_0^{\frac{\pi }{2}}=\frac{1}{2}\times \frac{{\pi }^2}{4}=\frac{{\pi }^2}{8}$$$$\Rightarrow \mathrm{I}=\frac{{\pi }^2}{8}....\mathrm{Trivial}.$$

Answered by Ar Brandon last updated on 28/Jul/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{arctanx}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}\mathrm{udu}={\left[\frac{{\mathrm{u}}^2}{2}\right]}_0^{\frac{\pi }{2}}=\frac{{\pi }^2}{8}$$

Commented by Ar Brandon last updated on 28/Jul/21

$$\mathrm{u}=\mathrm{arctanx}\Rightarrow \mathrm{du}=\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$

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