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
![F =16 ∫_0 ^∞ ((arctanx)/(1+x^2 ))dx =_(by parts) 16{ [arctan^2 x]_0 ^∞ −∫_0 ^∞ ((arctanx)/(1+x^2 ))dx} =16{(π^2 /4)−∫_0 ^∞ ((arctanx)/(1+x^2 ))dx}=4π^2 −F ⇒2F =4π^2 ⇒F =2π^2](Q134387.png)
$$\mathrm{F}\:=\mathrm{16}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{by}\:\mathrm{parts}} \:\:\:\mathrm{16}\left\{\:\left[\mathrm{arctan}^{\mathrm{2}} \mathrm{x}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right\} \\ $$$$=\mathrm{16}\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right\}=\mathrm{4}\pi^{\mathrm{2}} −\mathrm{F}\:\Rightarrow\mathrm{2F}\:=\mathrm{4}\pi^{\mathrm{2}} \:\Rightarrow\mathrm{F}\:=\mathrm{2}\pi^{\mathrm{2}} \\ $$