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Question Number 183157 by cortano1 last updated on 21/Dec/22

$$\int \frac{dx}{\sqrt[3]{x^3+2019}}=?$$

Answered by MJS_new last updated on 21/Dec/22

$$\int \frac{dx}{\sqrt[3]{x^3+a}}=$$$$[t=\frac{x}{\sqrt[3]{x^3+a}}\to dx=\frac{\sqrt[3]{{(x^3+a)}^4}}{a}]$$$$=\int \frac{dt}{1-t^3}=$$$$\left[\mathrm{the}\mathrm{usual}\mathrm{method}\right]$$$$=\frac{1}{6}\ln (t^2+t+1)+\frac{1}{3}\ln (t-1)+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t+1)}{3}$$$$\mathrm{now}\mathrm{insert}$$

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