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Question Number 126344 by mnjuly1970 last updated on 19/Dec/20

$$$$$$evaluate::{\int }_{-1}^0\frac{dx}{\sqrt[3]{1+x^3}}=?$$$$$$

Answered by Dwaipayan Shikari last updated on 19/Dec/20

$${\int }_0^1\frac{dt}{\sqrt[3]{1-t^3}}x=-t$$$$\Rightarrow \frac{1}{3}{\int }_0^1{u}^{-\frac{2}{3}}{(1-u)}^{-\frac{1}{3}}du{t}^3=u$$$$=\frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)=\frac{2\pi }{3\sqrt{3}}$$

Commented by mnjuly1970 last updated on 19/Dec/20

$$thanksalotmrpayan...$$

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