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Question Number 42085 by maxmathsup by imad last updated on 17/Aug/18
calculate ∫_1 ^(+∞) ((2x+1)/(3 +(x+1)^3 ))dx
$${calculate}\:\:\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}\:+\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}\: \\ $$
Commented by maxmathsup by imad last updated on 18/Aug/18
let A = ∫_1 ^(+∞) ((2x+1)/(3+(x+1)^3 )) dx changement x+1 =t give A = ∫_2 ^(+∞) ((2(t−1) +1)/(3 +t^3 )) dt = ∫_2 ^(+∞) ((2t−1)/(t^3 +3)) dt also changement^3 (√3)u =t give A = ∫_(2/3_(√3) ) ^(+∞) ((2(^3 (√3))u−1)/(3(1+u^3 )))^3 (√3)du let α =(2/((^3 (√3)))) = (2/3) (^3 (√3))^2 ∫_α ^(+∞) (u/(u^3 +1)) −^3 (√3) ∫_α ^(+∞) (du/(u^3 +1)) =(2/3)((2/α))^2 ∫_α ^(+∞) (u/(u^3 +1))du −(2/α) ∫_α ^(+∞) (du/(u^3 +1)) = ∫_α ^(+∞) (((8/(3α^2 ))u −(2/α))/(u^3 +1)) du =(2/α) ∫_α ^(+∞) (((4/(3α))u −1)/(u^3 +1)) du = (2/(3α^2 )) ∫_α ^(+∞) ((4u−3α)/(u^3 +1)) du let decompose F(u) =((4u−3α)/(u^3 +1)) F(u) =((4u−3α)/((u+1)(u^2 −u+1))) = (a/(u+1)) +((bu +c)/(u^2 −u +1)) a =lim_(u→−1) (u+1)F(u)=((−4−3α)/3) lim_(u→+∞) u F(u) =0 =a+b ⇒ b =((4+3α)/3) ⇒ F(u) =−((3α+4)/(3(u+1))) +(1/3) (((3α+4)u +3c)/(u^2 −u +1)) F(0) =−3α =−((3α+4)/3) +c ⇒c =−3α +((3α+4)/3) =((−6α +4)/3) ⇒ F(u) = −((3α+4)/(3(u+1))) +(1/3) (((3α+4)u −6α +4)/(u^2 −u +1)) ⇒ ∫_α ^(+∞) F(u)du =(−α−(4/3)) ∫_α ^(+∞) (du/(u+1)) +((3α+4)/6) ∫_α ^(+∞) ((2α−1+1)/(u^2 −u+1))du +((4−6α)/3) ∫_α ^(+∞) (du/(u^2 −u +1)) =[(−α−(4/3))ln∣u+1∣ +((3α+4)/6)ln(u^2 −u+1)]_0 ^(+∞) +((4−6α)/3) ∫_α ^(+∞) (du/(u^2 −u +1)) ...
$${let}\:\:{A}\:\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}+\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:{dx}\:\:{changement}\:\:{x}+\mathrm{1}\:={t}\:{give} \\ $$$${A}\:\:=\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\frac{\mathrm{2}\left({t}−\mathrm{1}\right)\:+\mathrm{1}}{\mathrm{3}\:+{t}^{\mathrm{3}} }\:{dt}\:=\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{3}} \:+\mathrm{3}}\:{dt}\:\:{also}\:{changement}\:^{\mathrm{3}} \sqrt{\mathrm{3}}{u}\:={t}\:{give} \\ $$$${A}\:=\:\:\int_{\frac{\mathrm{2}}{\mathrm{3}_{\sqrt{\mathrm{3}}} }} ^{+\infty} \:\:\:\:\:\:\frac{\mathrm{2}\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right){u}−\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{u}^{\mathrm{3}} \right)}\:^{\mathrm{3}} \sqrt{\mathrm{3}}{du}\:\:\:{let}\:\:\alpha\:=\frac{\mathrm{2}}{\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)} \\ $$$$=\:\:\frac{\mathrm{2}}{\mathrm{3}}\:\:\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)^{\mathrm{2}} \int_{\alpha} ^{+\infty} \:\:\:\frac{{u}}{{u}^{\mathrm{3}} \:+\mathrm{1}}\:\:−^{\mathrm{3}} \sqrt{\mathrm{3}}\:\:\:\int_{\alpha} ^{+\infty} \:\:\:\:\frac{{du}}{{u}^{\mathrm{3}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\alpha}\right)^{\mathrm{2}} \:\:\:\int_{\alpha} ^{+\infty} \:\:\frac{{u}}{{u}^{\mathrm{3}} \:+\mathrm{1}}{du}\:\:−\frac{\mathrm{2}}{\alpha}\:\:\int_{\alpha} ^{+\infty} \:\:\frac{{du}}{{u}^{\mathrm{3}} \:+\mathrm{1}} \\ $$$$=\:\int_{\alpha} ^{+\infty} \:\:\:\frac{\frac{\mathrm{8}}{\mathrm{3}\alpha^{\mathrm{2}} }{u}\:−\frac{\mathrm{2}}{\alpha}}{{u}^{\mathrm{3}} \:+\mathrm{1}}\:{du}\:\:=\frac{\mathrm{2}}{\alpha}\:\:\int_{\alpha} ^{+\infty} \:\:\:\:\frac{\frac{\mathrm{4}}{\mathrm{3}\alpha}{u}\:−\mathrm{1}}{{u}^{\mathrm{3}} \:+\mathrm{1}}\:{du} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}\alpha^{\mathrm{2}} }\:\int_{\alpha} ^{+\infty} \:\:\:\frac{\mathrm{4}{u}−\mathrm{3}\alpha}{{u}^{\mathrm{3}} \:+\mathrm{1}}\:{du}\:\:\:{let}\:{decompose}\:{F}\left({u}\right)\:=\frac{\mathrm{4}{u}−\mathrm{3}\alpha}{{u}^{\mathrm{3}} \:+\mathrm{1}} \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{4}{u}−\mathrm{3}\alpha}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)}\:=\:\frac{{a}}{{u}+\mathrm{1}}\:\:+\frac{{bu}\:+{c}}{{u}^{\mathrm{2}} −{u}\:+\mathrm{1}} \\ $$$${a}\:={lim}_{{u}\rightarrow−\mathrm{1}} \left({u}+\mathrm{1}\right){F}\left({u}\right)=\frac{−\mathrm{4}−\mathrm{3}\alpha}{\mathrm{3}} \\ $$$${lim}_{{u}\rightarrow+\infty} \:{u}\:{F}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow\:{b}\:=\frac{\mathrm{4}+\mathrm{3}\alpha}{\mathrm{3}}\:\:\Rightarrow \\ $$$${F}\left({u}\right)\:=−\frac{\mathrm{3}\alpha+\mathrm{4}}{\mathrm{3}\left({u}+\mathrm{1}\right)}\:\:+\frac{\mathrm{1}}{\mathrm{3}}\:\:\frac{\left(\mathrm{3}\alpha+\mathrm{4}\right){u}\:\:+\mathrm{3}{c}}{{u}^{\mathrm{2}} −{u}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=−\mathrm{3}\alpha\:\:=−\frac{\mathrm{3}\alpha+\mathrm{4}}{\mathrm{3}}\:\:+{c}\:\Rightarrow{c}\:=−\mathrm{3}\alpha\:+\frac{\mathrm{3}\alpha+\mathrm{4}}{\mathrm{3}}\:=\frac{−\mathrm{6}\alpha\:+\mathrm{4}}{\mathrm{3}}\:\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\:−\frac{\mathrm{3}\alpha+\mathrm{4}}{\mathrm{3}\left({u}+\mathrm{1}\right)}\:\:+\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\left(\mathrm{3}\alpha+\mathrm{4}\right){u}\:\:−\mathrm{6}\alpha\:+\mathrm{4}}{{u}^{\mathrm{2}} \:−{u}\:+\mathrm{1}}\:\:\Rightarrow \\ $$$$\int_{\alpha} ^{+\infty} {F}\left({u}\right){du}\:\:=\left(−\alpha−\frac{\mathrm{4}}{\mathrm{3}}\right)\:\int_{\alpha} ^{+\infty} \:\:\frac{{du}}{{u}+\mathrm{1}}\:\:\:+\frac{\mathrm{3}\alpha+\mathrm{4}}{\mathrm{6}}\:\:\int_{\alpha} ^{+\infty} \:\:\frac{\mathrm{2}\alpha−\mathrm{1}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du} \\ $$$$\:+\frac{\mathrm{4}−\mathrm{6}\alpha}{\mathrm{3}}\:\int_{\alpha} ^{+\infty} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} −{u}\:+\mathrm{1}} \\ $$$$=\left[\left(−\alpha−\frac{\mathrm{4}}{\mathrm{3}}\right){ln}\mid{u}+\mathrm{1}\mid\:+\frac{\mathrm{3}\alpha+\mathrm{4}}{\mathrm{6}}{ln}\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\mathrm{4}−\mathrm{6}\alpha}{\mathrm{3}}\:\int_{\alpha} ^{+\infty} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:−{u}\:+\mathrm{1}}\:… \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Aug/18
t=x+1 dt=dx ∫_2 ^∞ ((2(t−1)+1)/(3+t^3 ))dt ∫_2 ^∞ ((2t+1)/(3+t^3 ))dt ((2t+1)/(t^3 +(3^(1/3) )^3 ))=((2t+1)/((t+3^(1/3) )(t^2 −3^(1/3) t+3^(2/3) ))) let a=3^(1/3) ((2t+1)/((t+a)(t^2 −ta+a^2 )))=(P/(t+a))+((Qt+R)/((t^2 −ta+a^2 ))) 2t+1=P(t^2 −ta+a^2 )+(t+a)(Qt+R) 2t+1=t^2 (P)+t(−Pa)+(Pa^2 )+t^2 (Q)+t(R)+t(Qa)+aR 2t+1=t^2 (P+Q)+t(−Pa+R+Qa)+(Pa^2 +aR) P+Q=0 −Pa+Qa+R=2 Pa^2 +aR=1 Q=−P −Pa−Pa+((1−Pa^2 )/a)=2 −2Pa^2 +1−Pa^2 =2a −3Pa^2 =2a−1 P=((1−2a)/(3a^2 )) Q=((2a−1)/(3a^2 )) R=((1−Pa^2 )/a)=((1−((1−2a)/3))/a)=((3−1+2a)/(3a))=((2+2a)/(3a)) ∫_2 ^∞ (P/(t+a))+((Qt+R)/(t^2 −ta+a^2 )) dt P∫_2 ^∞ (dt/(t+a))+(Q/2)∫_2 ^∞ ((2t−a+a)/(t^2 −ta+a^2 ))+R∫_2 ^∞ (dt/(t^2 −2.t.(a/2)+(a^2 /4)+((3a^2 )/4))) P∫_2 ^∞ (dt/(t+a))+(Q/2)∫_2 ^∞ ((2t−a)/(t^2 −ta+a^2 ))dt+(((Qa)/2)+R)∫_2 ^∞ (dt/((t−(a/2))^2 +(((a(√3))/2))^2 )) Pln∣(t+a)∣_2 ^∞ +(Q/2)ln∣(t^2 −ta+a^2 )∣_2 ^∞ +(((Qa)/2)+R)×(2/(a(√3)))∣tan^(−1) (((t−(a/2))/((a(√3))/2)))∣_2 ^∞ to be clmpleted...
$${t}={x}+\mathrm{1}\:\:\:{dt}={dx} \\ $$$$\int_{\mathrm{2}} ^{\infty} \frac{\mathrm{2}\left({t}−\mathrm{1}\right)+\mathrm{1}}{\mathrm{3}+{t}^{\mathrm{3}} }{dt} \\ $$$$\int_{\mathrm{2}} ^{\infty} \frac{\mathrm{2}{t}+\mathrm{1}}{\mathrm{3}+{t}^{\mathrm{3}} }{dt} \\ $$$$\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{3}} +\left(\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} }=\frac{\mathrm{2}{t}+\mathrm{1}}{\left({t}+\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\left({t}^{\mathrm{2}} −\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} {t}+\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)} \\ $$$${let}\:{a}=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{2}{t}+\mathrm{1}}{\left({t}+{a}\right)\left({t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} \right)}=\frac{{P}}{{t}+{a}}+\frac{{Qt}+{R}}{\left({t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}{t}+\mathrm{1}={P}\left({t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} \right)+\left({t}+{a}\right)\left({Qt}+{R}\right) \\ $$$$\mathrm{2}{t}+\mathrm{1}={t}^{\mathrm{2}} \left({P}\right)+{t}\left(−{Pa}\right)+\left({Pa}^{\mathrm{2}} \right)+{t}^{\mathrm{2}} \left({Q}\right)+{t}\left({R}\right)+{t}\left({Qa}\right)+{aR} \\ $$$$\mathrm{2}{t}+\mathrm{1}={t}^{\mathrm{2}} \left({P}+{Q}\right)+{t}\left(−{Pa}+{R}+{Qa}\right)+\left({Pa}^{\mathrm{2}} +{aR}\right) \\ $$$${P}+{Q}=\mathrm{0} \\ $$$$−{Pa}+{Qa}+{R}=\mathrm{2} \\ $$$${Pa}^{\mathrm{2}} +{aR}=\mathrm{1} \\ $$$${Q}=−{P} \\ $$$$−{Pa}−{Pa}+\frac{\mathrm{1}−{Pa}^{\mathrm{2}} }{{a}}=\mathrm{2} \\ $$$$−\mathrm{2}{Pa}^{\mathrm{2}} +\mathrm{1}−{Pa}^{\mathrm{2}} =\mathrm{2}{a} \\ $$$$−\mathrm{3}{Pa}^{\mathrm{2}} =\mathrm{2}{a}−\mathrm{1} \\ $$$${P}=\frac{\mathrm{1}−\mathrm{2}{a}}{\mathrm{3}{a}^{\mathrm{2}} }\:\:\:\:\:{Q}=\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{3}{a}^{\mathrm{2}} } \\ $$$${R}=\frac{\mathrm{1}−{Pa}^{\mathrm{2}} }{{a}}=\frac{\mathrm{1}−\frac{\mathrm{1}−\mathrm{2}{a}}{\mathrm{3}}}{{a}}=\frac{\mathrm{3}−\mathrm{1}+\mathrm{2}{a}}{\mathrm{3}{a}}=\frac{\mathrm{2}+\mathrm{2}{a}}{\mathrm{3}{a}} \\ $$$$\int_{\mathrm{2}} ^{\infty} \:\frac{{P}}{{t}+{a}}+\frac{{Qt}+{R}}{{t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} }\:\:{dt} \\ $$$${P}\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{t}+{a}}+\frac{{Q}}{\mathrm{2}}\int_{\mathrm{2}} ^{\infty} \frac{\mathrm{2}{t}−{a}+{a}}{{t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} }+{R}\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}.{t}.\frac{{a}}{\mathrm{2}}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${P}\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{{t}+{a}}+\frac{{Q}}{\mathrm{2}}\int_{\mathrm{2}} ^{\infty} \frac{\mathrm{2}{t}−{a}}{{t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} }{dt}+\left(\frac{{Qa}}{\mathrm{2}}+{R}\right)\int_{\mathrm{2}} ^{\infty} \frac{{dt}}{\left({t}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${Pln}\mid\left({t}+{a}\right)\mid_{\mathrm{2}} ^{\infty} \:+\frac{{Q}}{\mathrm{2}}{ln}\mid\left({t}^{\mathrm{2}} −{ta}+{a}^{\mathrm{2}} \right)\mid_{\mathrm{2}} ^{\infty} +\left(\frac{{Qa}}{\mathrm{2}}+{R}\right)×\frac{\mathrm{2}}{{a}\sqrt{\mathrm{3}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{{a}}{\mathrm{2}}}{\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\mid_{\mathrm{2}} ^{\infty} \\ $$$${to}\:{be}\:{clmpleted}… \\ $$$$ \\ $$$$ \\ $$

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