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Question Number 137876 by Bekzod Jumayev last updated on 07/Apr/21

Answered by MJS_new last updated on 07/Apr/21

$$\int \frac{dx}{{(x^3-1)}^{1/3}}=$$$$[t=\frac{x}{{(x^3-1)}^{1/3}}\to dx=-{(x^3-1)}^{4/3}]$$$$=\int \frac{dt}{1-t^3}=\int \frac{dt}{(1-t)(1+t+t^2)}=$$$$=\frac{1}{3}\int \frac{dt}{1-t}+\frac{1}{6}\int \frac{2t+1}{t^2+t+1}dt+\frac{1}{2}\int \frac{dt}{t^2+t+1}=$$$$=-\frac{1}{3}\ln (1-t)+\frac{1}{6}\ln (t^2+t+1)+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t+1)}{3}=$$$$=\frac{1}{6}\ln \frac{t^2+t+1}{{(t-1)}^2}+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t+1)}{3}$$$$\mathrm{the}\mathrm{borders}:2\to \frac{2}{7^{1/3}};+\mathrm{\infty }\to 1$$$$\Rightarrow \mathrm{integral}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{converge}$$

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