Question Number 42796 by maxmathsup by imad last updated on 02/Sep/18
$${calculate}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{arctan}\left({x}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 04/Sep/18
![we have I = ∫_0 ^1 ((1+x^2 −1)/(1+x^2 )) arctan(x)dx= ∫_0 ^1 arctanx −∫_0 ^1 ((arctanx)/(1+x^2 ))dx but by parts ∫_0 ^1 arctanxdx =[x arctanx]_0 ^1 −∫_0 ^1 (x/(1+x^2 ))dx =(π/4) −[(1/2)ln(1+x^2 )]_0 ^1 =(π/4) −((ln(2))/2) also by parts u^′ =(1/(1+x^2 )) and v=arctan(x) ∫_0 ^1 ((arctan(x))/(1+x^2 )) dx = [arctan^2 x]_0 ^1 −∫_0 ^1 ((arctanx)/(1+x^2 )) dx ⇒ 2 ∫_0 ^1 ((arctan(x))/(1+x^2 ))dx =(π^2 /(16)) ⇒∫_0 ^1 ((arctan(x))/(1+x^2 ))dx =(π^2 /(32)) ⇒ I =(π/4) −((ln(2))/2) −(π^2 /(32)) .](Q42917.png)
$${we}\:{have}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{arctan}\left({x}\right){dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctanx}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{but} \\ $$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctanxdx}\:=\left[{x}\:{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\pi}{\mathrm{4}}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:{also}\:{by}\:{parts}\:{u}^{'} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{and}\:{v}={arctan}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=\:\:\left[{arctan}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctanx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\Rightarrow \\ $$$$\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\:\Rightarrow \\ $$$${I}\:\:=\frac{\pi}{\mathrm{4}}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
$${t}={tan}^{−\mathrm{1}} {x}\:\:\:{dt}=\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \frac{{tan}^{\mathrm{2}} {t}×{t}}{}{dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \left({sec}^{\mathrm{2}} {t}−\mathrm{1}\right){t}\:{dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} {tsec}^{\mathrm{2}} {tdt}−\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} {tdt} \\ $$$$\mid{ttant}−{lnsect}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} \\ $$$$\frac{\Pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}−\frac{\prod^{\mathrm{2}} }{\mathrm{32}} \\ $$