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Question Number 159292 by HongKing last updated on 15/Nov/21
Find:  𝛀 =∫_( 0) ^( ∞) ((x arctan(x))/((x + 1)(x^2  + 1))) dx
$$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\frac{\mathrm{x}\:\mathrm{arctan}\left(\mathrm{x}\right)}{\left(\mathrm{x}\:+\:\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)}\:\mathrm{dx} \\ $$$$ \\ $$
Answered by mindispower last updated on 15/Nov/21
=∫_0 ^∞ ((arctan(x))/(1+x^2 ))βˆ’((arctan(x))/((1+x)(1+x^2 )))dx  =(Ο€^2 /8)βˆ’βˆ«_0 ^(Ο€/2) ((xcos(x))/(cos(x)+sin(x)))dx∣_(=A)   B=∫_0 ^(Ο€/2) ((xsin(x))/(sin(x)+cos(x)))  A+B=(Ο€^2 /8)  Aβˆ’B=∫_0 ^(Ο€/2) ((x(cos(x)βˆ’sin(x)))/(cos(x)+sin(x)))dx  =[xln(cos(x)+sin(x))]_0 ^(Ο€/2) βˆ’βˆ«_0 ^(Ο€/2) ln(cos(x)+sin(x){dx  =βˆ’βˆ«_0 ^(Ο€/2) ln((√2)sin(x+(Ο€/4)))d=βˆ’(Ο€/2)ln((√2))βˆ’βˆ«_(Ο€/4) ^((3Ο€)/4) ln(sin(x))dx  =βˆ’βˆ«_(Ο€/4) ^(Ο€/2) ln(sin(x))dxβˆ’βˆ«_0 ^(Ο€/4) lncos(x)  =βˆ’2∫_0 ^(Ο€/4) ln(cos(x))dx=βˆ’2.(1/4)(2Gβˆ’Ο€ln(2))  catalan constant  A=(1/2)(Aβˆ’B+A+B)=(1/2)((Ο€^2 /8)βˆ’(Ο€/2)ln((√2))βˆ’G+(Ο€/2)ln(2))  =(Ο€^2 /(16))+(Ο€/8)ln(2)βˆ’(G/2)  Ξ©=(Ο€^2 /(16))+(G/2)βˆ’((Ο€ln(2))/8)
$$=\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }βˆ’\frac{{arctan}\left({x}\right)}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xcos}\left({x}\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx}\mid_{={A}} \\ $$$${B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xsin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)} \\ $$$${A}+{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${A}βˆ’{B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\left({cos}\left({x}\right)βˆ’{sin}\left({x}\right)\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx} \\ $$$$=\left[{xln}\left({cos}\left({x}\right)+{sin}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\left({x}\right)+{sin}\left({x}\right)\left\{{dx}\right.\right. \\ $$$$=βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){d}=βˆ’\frac{\pi}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\right)βˆ’\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} {ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$=βˆ’\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {lncos}\left({x}\right) \\ $$$$=βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right){dx}=βˆ’\mathrm{2}.\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{G}βˆ’\pi{ln}\left(\mathrm{2}\right)\right) \\ $$$${catalan}\:{constant} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left({A}βˆ’{B}+{A}+{B}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}βˆ’\frac{\pi}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\right)βˆ’{G}+\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)βˆ’\frac{{G}}{\mathrm{2}} \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{{G}}{\mathrm{2}}βˆ’\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$ \\ $$
Commented by HongKing last updated on 15/Nov/21
cool my dear Ser thank you so much
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by mindispower last updated on 15/Nov/21
you are welcom  have a nice day
$${you}\:{are}\:{welcom} \\ $$$${have}\:{a}\:{nice}\:{day} \\ $$$$ \\ $$
Commented by HongKing last updated on 15/Nov/21
thank you very much my dear Ser
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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