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Question Number 135633 by metamorfose last updated on 14/Mar/21

∫(1/(x^(1/3) +1))dx=...?

$$\int\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}{dx}=...? \\ $$

Answered by Ñï= last updated on 14/Mar/21

∫(dx/(x^(1/3) +1))=^(t=x^(1/3) ) ∫((3t^2 dt)/(t+1))=3∫(((t+1)(t−1)+1)/(t+1))dt=3∫{(t−1)+(1/(t+1))}dt  =(3/2)t^2 −3t+3ln(t+1)+C=(3/2)x^(2/3) −3x^(1/3) +3ln(1+x^(1/3) )+C

$$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}\overset{{t}={x}^{\frac{\mathrm{1}}{\mathrm{3}}} } {=}\int\frac{\mathrm{3}{t}^{\mathrm{2}} {dt}}{{t}+\mathrm{1}}=\mathrm{3}\int\frac{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)+\mathrm{1}}{{t}+\mathrm{1}}{dt}=\mathrm{3}\int\left\{\left({t}−\mathrm{1}\right)+\frac{\mathrm{1}}{{t}+\mathrm{1}}\right\}{dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{3}{ln}\left({t}+\mathrm{1}\right)+{C}=\frac{\mathrm{3}}{\mathrm{2}}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{3}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}{ln}\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)+{C} \\ $$

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