Question Number 137876 by Bekzod Jumayev last updated on 07/Apr/21
Answered by MJS_new last updated on 07/Apr/21
![∫(dx/((x^3 −1)^(1/3) ))= [t=(x/((x^3 −1)^(1/3) )) → dx=−(x^3 −1)^(4/3) ] =∫(dt/(1−t^3 ))=∫(dt/((1−t)(1+t+t^2 )))= =(1/3)∫(dt/(1−t))+(1/6)∫((2t+1)/(t^2 +t+1))dt+(1/2)∫(dt/(t^2 +t+1))= =−(1/3)ln (1−t) +(1/6)ln (t^2 +t+1) +((√3)/3)arctan (((√3)(2t+1))/3) = =(1/6)ln ((t^2 +t+1)/((t−1)^2 )) +((√3)/3)arctan (((√3)(2t+1))/3) the borders: 2 → (2/7^(1/3) ) ; +∞ → 1 ⇒ integral doesn′t converge](https://www.tinkutara.com/question/Q137888.png)
$$\int\frac{{dx}}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} }= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} }\:\rightarrow\:{dx}=−\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} \right] \\ $$$$=\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{3}} }=\int\frac{{dt}}{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{dt}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{1}−{t}\right)\:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{3}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\mathrm{the}\:\mathrm{borders}:\:\mathrm{2}\:\rightarrow\:\frac{\mathrm{2}}{\mathrm{7}^{\mathrm{1}/\mathrm{3}} }\:;\:+\infty\:\rightarrow\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{integral}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{converge} \\ $$