Find-0-x-arctan-x-x-1-x-2-1-dx-

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 ))−((arctan(x))/((1+x)(1+x^2 )))dx =(π^2 /8)−∫_0 ^(π/2) ((xcos(x))/(cos(x)+sin(x)))dx∣_(=A) B=∫_0 ^(π/2) ((xsin(x))/(sin(x)+cos(x))) A+B=(π^2 /8) A−B=∫_0 ^(π/2) ((x(cos(x)−sin(x)))/(cos(x)+sin(x)))dx =[xln(cos(x)+sin(x))]_0 ^(π/2) −∫_0 ^(π/2) ln(cos(x)+sin(x){dx =−∫_0 ^(π/2) ln((√2)sin(x+(π/4)))d=−(π/2)ln((√2))−∫_(π/4) ^((3π)/4) ln(sin(x))dx =−∫_(π/4) ^(π/2) ln(sin(x))dx−∫_0 ^(π/4) lncos(x) =−2∫_0 ^(π/4) ln(cos(x))dx=−2.(1/4)(2G−πln(2)) catalan constant A=(1/2)(A−B+A+B)=(1/2)((π^2 /8)−(π/2)ln((√2))−G+(π/2)ln(2)) =(π^2 /(16))+(π/8)ln(2)−(G/2) Ω=(π^2 /(16))+(G/2)−((πln(2))/8)

$$=\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{{arctan}\left({x}\right)}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xcos}\left({x}\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx}\mid_{={A}} \\ $$$${B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xsin}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)} \\ $$$${A}+{B}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${A}−{B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\left({cos}\left({x}\right)−{sin}\left({x}\right)\right)}{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx} \\ $$$$=\left[{xln}\left({cos}\left({x}\right)+{sin}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\left({x}\right)+{sin}\left({x}\right)\left\{{dx}\right.\right. \\ $$$$=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){d}=−\frac{\pi}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\right)−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} {ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$=−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {lncos}\left({x}\right) \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right){dx}=−\mathrm{2}.\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{G}−\pi{ln}\left(\mathrm{2}\right)\right) \\ $$$${catalan}\:{constant} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left({A}−{B}+{A}+{B}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\pi}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\right)−{G}+\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}} \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{{G}}{\mathrm{2}}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$ \\ $$

Commented by HongKing last updated on 15/Nov/21