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Question Number 117034 by Lordose last updated on 09/Oct/20

$$\int \frac{\mathbf{dx}}{\sqrt[3]{1+{\mathbf{x}}^3}}$$

Answered by MJS_new last updated on 09/Oct/20

$$\int \frac{dx}{\sqrt[3]{x^3+1}}=$$$$[t=\frac{\sqrt[3]{x^3+1}}{x}\to dx=-x^2\sqrt[3]{{(x^3+1)}^2}]$$$$=-\int \frac{t}{t^3-1}dt=\frac{1}{3}\int \frac{t-1}{t^2+t+1}-\frac{1}{3}\int \frac{dt}{t-1}=$$$$=\frac{1}{6}\ln (t^2+t+1)-\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t+1)}{3}-\frac{1}{3}\ln (t-1)$$$$\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{easy}$$

Commented by Lordose last updated on 09/Oct/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by bobhans last updated on 09/Oct/20

$$\mathrm{integral}\mathrm{lover}$$

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