Question Number 117034 by Lordose last updated on 09/Oct/20
$$\int\frac{\boldsymbol{\mathrm{dx}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{3}} }} \\ $$
Answered by MJS_new last updated on 09/Oct/20
![∫(dx/( ((x^3 +1))^(1/3) ))= [t=(((x^3 +1))^(1/3) /x) → dx=−x^2 (((x^3 +1)^2 ))^(1/3) ] =−∫(t/(t^3 −1))dt=(1/3)∫((t−1)/(t^2 +t+1))−(1/3)∫(dt/(t−1))= =(1/6)ln (t^2 +t+1) −((√3)/3)arctan (((√3)(2t+1))/3) −(1/3)ln (t−1) the rest is easy](Q117040.png)
$$\int\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}}{{x}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }\right] \\ $$$$=−\int\frac{{t}}{{t}^{\mathrm{3}} −\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}−\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left({t}−\mathrm{1}\right) \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$
Commented by Lordose last updated on 09/Oct/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by bobhans last updated on 09/Oct/20
$$\mathrm{integral}\:\mathrm{lover} \\ $$