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Integration Questions

Question Number 134303 by bramlexs22 last updated on 02/Mar/21

$$\mathcal{F}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{16}\:\mathrm{arctan}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\$$

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

$$\mathcal{F}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{16}\:\mathrm{arctan}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\$$$$=\mathrm{16}\int_{\mathrm{0}} ^{\infty} \mathrm{tan}^{−\mathrm{1}} {xdtan}^{−\mathrm{1}} {x} \\$$$$=\mathrm{8}\left({tan}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} \mid_{\mathrm{0}} ^{\infty} \\$$$$=\mathrm{8}×\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \\$$$$=\mathrm{2}\pi^{\mathrm{2}} \\$$

Answered by mathmax by abdo last updated on 02/Mar/21

$$\mathrm{F}\:=\mathrm{16}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{by}\:\mathrm{parts}} \:\:\:\mathrm{16}\left\{\:\left[\mathrm{arctan}^{\mathrm{2}} \mathrm{x}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right\} \\$$$$=\mathrm{16}\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right\}=\mathrm{4}\pi^{\mathrm{2}} −\mathrm{F}\:\Rightarrow\mathrm{2F}\:=\mathrm{4}\pi^{\mathrm{2}} \:\Rightarrow\mathrm{F}\:=\mathrm{2}\pi^{\mathrm{2}} \\$$