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Question Number 41302 by math khazana by abdo last updated on 05/Aug/18

calculate ∫_1 ^(+∞) (dx/(x^2 (√(x^2 +x+1))))

$${calculate}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}} \\ $$

Commented by maxmathsup by imad last updated on 05/Aug/18

let I = ∫_1 ^(+∞) (dx/(x^2 (√(x^2 +x+1)))) I = ∫_1 ^(+∞) (dx/(x^2 (√((x+(1/2))^2 +(3/4))))) changement x+(1/2)=((√3)/2)tanθ give I = ∫_(arctan((1/(√3)))) ^(π/2) (1/((((√3)/2)tanθ−(1/2))^2 ((√3)/2))) cos^2 θ ((√3)/2)(1+tan^2 θ)dθ = ∫_(π/6) ^(π/2) ((4dθ)/(((√3)tanθ−1)^2 )) = ∫_(π/6) ^(π/2) (4/(3tan^2 θ −2(√3)tanθ +1))dθ =_(tanθ =x) ∫_(1/(√3)) ^(+∞) (4/((3x^2 −2(√(3x))+1))) (dx/(1+x^2 )) let decompose F(x) = (4/((x^2 +1)(3x^2 −2(√3)x+1))) =(4/(((√3)x−1)^2 (x^2 +1))) F(x)=(a/((√3)x−1)) +(b/(((√3)x−1)^2 )) +((bx +c)/(x^2 +1)) b=lim_(x→(1/(√3))) ((√3)x−1)^2 F(x)= (4/(4/3)) =3 lim_(x→+∞) xF(x)=0 =(a/(√3)) +b ⇒b=−(a/(√3)) ⇒ F(x)= (a/((√3)x−1)) +(3/(((√3)x−1)^2 )) +((−(a/(√3))x +c)/(x^2 +1)) F(0) =−a +3 +c ⇒c=a−3 ⇒ F(x)= (a/((√3)x−1)) +(3/(((√3)x−1)^2 )) +((−(a/(√3))x +a−3)/(x^2 +1)) F((√3)) = (4/(4(4))) =(1/4) =(a/2) +(3/4) +((−3)/4) ⇒(a/2) =(1/4) ⇒a=(1/2) ⇒ F(x) = (1/(2((√3)x−1))) +(3/(((√3)x−1)^2 )) +((−(1/(2(√3)))x −(5/2))/(x^2 +1)) ⇒ ∫ F(x)dx =(1/(2(√3))) ∫ (dx/(x−(1/(√3)))) + ∫ (dx/((x−(1/(√3)))^2 )) −(1/(2(√3))) ∫ ((x+5(√3))/(x^2 +1)) dx =(1/(2(√3)))ln∣x−(1/(√3))∣ −(1/(x−(1/(√3)))) −(1/(4(√3))) ∫ ((2x)/(x^2 +1)) −(5/2) ∫ (dx/(x^2 +1)) =(1/(2(√3)))ln∣x−(1/(√3))∣ −(1/(x−(1/(√3)))) −(1/(4(√3)))ln(x^2 +1)−(5/2) arctanx +c ⇒ ∫_(1/(√3)) ^(+∞) F(x)dx =....(put the lmits)

$${let}\:{I}\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}\:\:{changement}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}\theta\:{give} \\ $$$${I}\:=\:\int_{{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}\theta−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:{cos}^{\mathrm{2}} \theta\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}{d}\theta}{\left(\sqrt{\mathrm{3}}{tan}\theta−\mathrm{1}\right)^{\mathrm{2}} }\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{4}}{\mathrm{3}{tan}^{\mathrm{2}} \theta\:−\mathrm{2}\sqrt{\mathrm{3}}{tan}\theta\:+\mathrm{1}}{d}\theta \\ $$$$=_{{tan}\theta\:={x}} \:\:\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{+\infty} \:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}{x}}+\mathrm{1}\right)}\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{let}\:{decompose}\: \\ $$$${F}\left({x}\right)\:=\:\:\frac{\mathrm{4}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}\:=\frac{\mathrm{4}}{\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{\sqrt{\mathrm{3}}{x}−\mathrm{1}}\:+\frac{{b}}{\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{bx}\:+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${b}={lim}_{{x}\rightarrow\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\frac{\mathrm{4}}{\frac{\mathrm{4}}{\mathrm{3}}}\:=\mathrm{3} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:=\frac{{a}}{\sqrt{\mathrm{3}}}\:+{b}\:\Rightarrow{b}=−\frac{{a}}{\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{\sqrt{\mathrm{3}}{x}−\mathrm{1}}\:+\frac{\mathrm{3}}{\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−\frac{{a}}{\sqrt{\mathrm{3}}}{x}\:+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=−{a}\:+\mathrm{3}\:+{c}\:\Rightarrow{c}={a}−\mathrm{3}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{\sqrt{\mathrm{3}}{x}−\mathrm{1}}\:+\frac{\mathrm{3}}{\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−\frac{{a}}{\sqrt{\mathrm{3}}}{x}\:+{a}−\mathrm{3}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\sqrt{\mathrm{3}}\right)\:=\:\frac{\mathrm{4}}{\mathrm{4}\left(\mathrm{4}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\:=\frac{{a}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{4}}\:+\frac{−\mathrm{3}}{\mathrm{4}}\:\Rightarrow\frac{{a}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)}\:+\frac{\mathrm{3}}{\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{x}\:−\frac{\mathrm{5}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\int\:\:\:\:\frac{{dx}}{{x}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\:+\:\int\:\:\:\:\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\:\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\int\:\:\:\frac{{x}+\mathrm{5}\sqrt{\mathrm{3}}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid{x}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\mid\:−\frac{\mathrm{1}}{{x}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\:\int\:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{5}}{\mathrm{2}}\:\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\mid{x}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\mid\:−\frac{\mathrm{1}}{{x}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)−\frac{\mathrm{5}}{\mathrm{2}}\:{arctanx}\:+{c}\:\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{+\infty} {F}\left({x}\right){dx}\:=....\left({put}\:{the}\:{lmits}\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

∫_1 ^∞ ((1/x^2 )/(√(x^2 (1+(1/x)+(1/x^(2 ) )))))dx t=(1/x) dt=−(1/x^2 )dx ∫_1 ^0 ((−dt)/(√((1/t^2 )(1+t+t^2 )))) ∫_0 ^1 ((tdt)/(√(t^2 +t+1))) =(1/2)∫_0 ^1 ((2t+1−1)/(√(t^2 +t+1)))dt =(1/2)∫_0 ^1 ((2t+1)/(√(t^2 +t+1)))−(1/2)∫_0 ^1 (dt/(√(t^2 +2.t.(1/2)+(1/4)+1−(1/4)))) =(1/2)∫_0 ^1 ((d(t^2 +t+1))/(√(t^2 +t+1)))−(1/2)∫_0 ^1 (dt/(√((t+(1/2))^2 +(((√3)/2))^2 ))) =(1/2)∣((√(t^2 +t+1))/(1/2))∣_0 ^1 −(1/2)∣ln{(t+(1/2))+(√((t+(1/2))^2 +(((√3)/2))^2 )) }∣_0 ^1 =((√3) −1)−(1/2){ln((3/2)+(√((3/4)+(3/4))) )−ln((1/2)+1)} =((√3) −1)−(1/2){ln(((3+(√6))/2))−ln((3/2))} =((√3) −1)−(1/2){ln(((3+(√6))/3))}

$$\int_{\mathrm{1}} ^{\infty} \frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} }\right)}}{dx} \\ $$$${t}=\frac{\mathrm{1}}{{x}}\:\:\:{dt}=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{0}} \frac{−{dt}}{\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tdt}}{\sqrt{{t}^{\mathrm{2}} +{t}+\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}+\mathrm{1}−\mathrm{1}}{\sqrt{{t}^{\mathrm{2}} +{t}+\mathrm{1}}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}+\mathrm{1}}{\sqrt{{t}^{\mathrm{2}} +{t}+\mathrm{1}}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\sqrt{{t}^{\mathrm{2}} +\mathrm{2}.{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{\sqrt{{t}^{\mathrm{2}} +{t}+\mathrm{1}}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{\sqrt{{t}^{\mathrm{2}} +{t}+\mathrm{1}}}{\frac{\mathrm{1}}{\mathrm{2}}}\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\mid{ln}\left\{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right\}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}\:\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\right\} \\ $$$$=\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{3}+\sqrt{\mathrm{6}}}{\mathrm{2}}\right)−{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right\} \\ $$$$=\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{3}+\sqrt{\mathrm{6}}}{\mathrm{3}}\right)\right\} \\ $$

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