Evaluate-the-integral-x-1-3-1-x-1-3-1-dx-

Question Number 122689 by john santu last updated on 19/Nov/20

$$\:{Evaluate}\:{the}\:{integral}\: \\ $$$$\:\:\int\:\frac{\sqrt[{\mathrm{3}}]{{x}}\:+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{1}}\:{dx}\: \\ $$

Answered by bobhans last updated on 19/Nov/20

let x = u^3 ⇒dx = 3u^2 du C(x)=∫((u+1)/(u−1)).(3u^2 du ) C(x) = 3∫ ((u^3 +u^2 )/(u−1)) du C(x)= 3 ∫ (u^2 +2u+2+(2/(u−1))) du C(x)=3((1/3)u^3 +u^2 +2u+2ln ∣u−1∣)+c C(x)=x+3(x^2 )^(1/3) +6(x)^(1/3) +6ln ∣(x)^(1/3) −1∣ + c

$$\:{let}\:{x}\:=\:{u}^{\mathrm{3}} \:\Rightarrow{dx}\:=\:\mathrm{3}{u}^{\mathrm{2}} \:{du}\: \\ $$$${C}\left({x}\right)=\int\frac{{u}+\mathrm{1}}{{u}−\mathrm{1}}.\left(\mathrm{3}{u}^{\mathrm{2}} \:{du}\:\right)\: \\ $$$${C}\left({x}\right)\:=\:\mathrm{3}\int\:\frac{{u}^{\mathrm{3}} +{u}^{\mathrm{2}} }{{u}−\mathrm{1}}\:{du}\: \\ $$$${C}\left({x}\right)=\:\mathrm{3}\:\int\:\left({u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{2}+\frac{\mathrm{2}}{{u}−\mathrm{1}}\right)\:{du} \\ $$$${C}\left({x}\right)=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} +{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{2ln}\:\mid{u}−\mathrm{1}\mid\right)+{c} \\ $$$${C}\left({x}\right)={x}+\mathrm{3}\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\:+\mathrm{6}\sqrt[{\mathrm{3}}]{{x}}\:+\mathrm{6ln}\:\mid\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{1}\mid\:+\:{c} \\ $$

Commented by MJS_new last updated on 19/Nov/20

another possibility: ∫((x^(1/3) +1)/(x^(1/3) −1))dx=∫dx+∫(2/(x^(1/3) −1))dx [t=x^(1/3) −1 → dx=3x^(2/3) dt] =x+6∫(((t+1)^2 )/t)dt=x+3t(t+4)+6ln t = =x+3(x^(1/3) −1)(x^(1/3) +3)+6ln ∣x^(1/3) −1∣ +C

$$\mathrm{another}\:\mathrm{possibility}: \\ $$$$\int\frac{{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{dx}=\int{dx}+\int\frac{\mathrm{2}}{{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{dx} \\ $$$$\:\:\:\:\:\left[{t}={x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}\:\rightarrow\:{dx}=\mathrm{3}{x}^{\mathrm{2}/\mathrm{3}} {dt}\right] \\ $$$$={x}+\mathrm{6}\int\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}}{dt}={x}+\mathrm{3}{t}\left({t}+\mathrm{4}\right)+\mathrm{6ln}\:{t}\:= \\ $$$$={x}+\mathrm{3}\left({x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}\right)\left({x}^{\mathrm{1}/\mathrm{3}} +\mathrm{3}\right)+\mathrm{6ln}\:\mid{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}\mid\:+{C} \\ $$$$ \\ $$

Commented by john santu last updated on 19/Nov/20

$${thank}\:{you} \\ $$

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