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Given the integral: \( \displaystyle I=\int_{0}^{1}\frac{dx}{\sqrt{1-x^3}} \).


a) Prove that the integrand does not have an elementary antiderivative.
b) Evaluate \( I \).

Making the substitution \(y = x^3\) we obtain \[ I = \frac{1}{3} \int_0^1 y^{-2/3}(1-y)^{-1/2} \, dy.\] It is well known that \[ B(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}.\] So \(I = \frac{\Gamma(1/3)\Gamma(1/2)}{3\Gamma(5/6)}.\)

A theorem of Chebyshev says that \(x^m(1-x)^n\) has an elementary antiderivative if and only if at least one of \(m,n,m+n\) is an integer. So in our situation this is not the case and the intergrand has no elementary antiderivative.