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Question Number 43904 by abdo.msup.com last updated on 17/Sep/18
calculate ∫_0 ^∞ (((1+x)^(1/3) −1)/(x(1+x)^(2/3) ))dx
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}}{{x}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 19/Sep/18
changement (1+x)^(1/3) =t give 1+x =t^3 ⇒ I =∫_1 ^(+∞) ((t−1)/((t^3 −1)t^2 )) (3t^2 )dt = ∫_1 ^(+∞) (3/(t^2 +t +1))dt = ∫_1 ^(+∞) ((3dt)/((t+(1/2))^2 +(3/4))) =_(t+(1/2)=((√3)/2)u) (4/3) ∫_(√3) ^(+∞) (3/((1+u^2 )))((√3)/2)du =2(√3) ∫_(√3) ^(+∞) (du/(1+u^2 )) =2(√3)[arctan(u)]_(√3) ^(+∞) =2(√3){(π/2)−(π/3)}=2(√3).(π/6) I = ((π(√3))/3) .
$${changement}\:\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:={t}\:{give}\:\mathrm{1}+{x}\:={t}^{\mathrm{3}} \:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}−\mathrm{1}}{\left({t}^{\mathrm{3}} −\mathrm{1}\right){t}^{\mathrm{2}} }\:\:\left(\mathrm{3}{t}^{\mathrm{2}} \right){dt}\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\mathrm{3}}{{t}^{\mathrm{2}} \:+{t}\:+\mathrm{1}}{dt} \\ $$$$=\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{\mathrm{3}{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \:\:\frac{\mathrm{4}}{\mathrm{3}}\:\int_{\sqrt{\mathrm{3}}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{3}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$=\mathrm{2}\sqrt{\mathrm{3}}\:\int_{\sqrt{\mathrm{3}}} ^{+\infty} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\sqrt{\mathrm{3}}\left[{arctan}\left({u}\right)\right]_{\sqrt{\mathrm{3}}} ^{+\infty} \:=\mathrm{2}\sqrt{\mathrm{3}}\left\{\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right\}=\mathrm{2}\sqrt{\mathrm{3}}.\frac{\pi}{\mathrm{6}} \\ $$$${I}\:=\:\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18
1+x=t^3 dx=3t^2 dt ∫_1 ^∞ ((t−1)/((t^3 −1)(t^2 )))dt ∫_1 ^∞ (dt/(t^2 (t^2 +t+1))) (1/(t^2 (t^2 +t+1)))=(a/t)+(b/t^2 )+((ct+d)/(t^2 +t+1)) 1=at(t^2 +t+1)+b(t^2 +t+1)+(ct+d)t^2 1=a(t^3 +t^2 +t)+b(t^2 +t+1)+(ct^3 +dt^2 ) 1=t^3 (a+c)+t^2 (a+b+d)+t(a+b)+b b=1 a+c=0 a+1+d=0 a+1=0 a=−1 c=1 −1+1+d=0 d=0 a=−1 b=1 c=1 d=0 ∫_1 ^∞ (dt/(t^2 (t^2 +t+1)))=∫_1 ^∞ ((−1)/t)dt+∫_1 ^∞ (1/t^2 )dt+∫_1 ^∞ (t/(t^2 +t+1))dt =∣−lnt∣_1 ^∞ +∣(((−1)/t))∣_1 ^∞ +(1/2)∫_1 ^∞ ((2t+1−1)/(t^2 +t+1))dt =(−ln∞)+1+(1/2)∫_1 ^∞ ((d(t^2 +t+1))/(t^2 +t+1))−(1/2)∫_1 ^∞ (dt/(t^2 +2.t.(1/2)+(1/4)+1−(1/4))) =(−ln∞)+1+(1/2)∣ln(t^2 +t+1)∣_1 ^∞ −(1/2)∫_1 ^∞ (dt/((t+(1/2))^2 +((((√3)[)/2))^2 )) =(−ln∞)+1+(1/2)ln(∞)+(1/2)ln3−(1/2)×(1/((√3)/2))∣tan^(−1) (((t+(1/2))/((√3)/2)))∣_1 ^∞ =(−(1/2)ln∞)+1+(1/2)ln3−(1/( (√3)))((Π/2)−(Π/3)) =(−(1/2)ln∞)+1+(1/2)ln3−(1/( (√3)))((Π/6))
$$\mathrm{1}+{x}={t}^{\mathrm{3}} \:\:\:\:{dx}=\mathrm{3}{t}^{\mathrm{2}} {dt} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{t}−\mathrm{1}}{\left({t}^{\mathrm{3}} −\mathrm{1}\right)\left({t}^{\mathrm{2}} \right)}{dt} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}=\frac{{a}}{{t}}+\frac{{b}}{{t}^{\mathrm{2}} }+\frac{{ct}+{d}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}} \\ $$$$\mathrm{1}={at}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)+{b}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)+\left({ct}+{d}\right){t}^{\mathrm{2}} \\ $$$$\mathrm{1}={a}\left({t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}\right)+{b}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)+\left({ct}^{\mathrm{3}} +{dt}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}={t}^{\mathrm{3}} \left({a}+{c}\right)+{t}^{\mathrm{2}} \left({a}+{b}+{d}\right)+{t}\left({a}+{b}\right)+{b} \\ $$$${b}=\mathrm{1} \\ $$$${a}+{c}=\mathrm{0} \\ $$$${a}+\mathrm{1}+{d}=\mathrm{0} \\ $$$${a}+\mathrm{1}=\mathrm{0}\:\:\:{a}=−\mathrm{1} \\ $$$${c}=\mathrm{1} \\ $$$$−\mathrm{1}+\mathrm{1}+{d}=\mathrm{0}\:\:\:{d}=\mathrm{0} \\ $$$${a}=−\mathrm{1}\:\:\:{b}=\mathrm{1}\:\:\:{c}=\mathrm{1}\:\:\:{d}=\mathrm{0} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}=\int_{\mathrm{1}} ^{\infty} \frac{−\mathrm{1}}{{t}}{dt}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt}+\int_{\mathrm{1}} ^{\infty} \frac{{t}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt} \\ $$$$=\mid−{lnt}\mid_{\mathrm{1}} ^{\infty} +\mid\left(\frac{−\mathrm{1}}{{t}}\right)\mid_{\mathrm{1}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{t}+\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt} \\ $$$$=\left(−{ln}\infty\right)+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{{d}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}.{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\left(−{ln}\infty\right)+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mid{ln}\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)\mid_{\mathrm{1}} ^{\infty} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}\left[\right.}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\left(−{ln}\infty\right)+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\infty\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\mid_{\mathrm{1}} ^{\infty} \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\infty\right)+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\frac{\Pi}{\mathrm{2}}−\frac{\Pi}{\mathrm{3}}\right) \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\infty\right)+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\frac{\Pi}{\mathrm{6}}\right) \\ $$

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