Evaluate the integral I=∫01dxx3+1I = \int_0^1 \frac{dx}{\sqrt{x^3+1}}I=∫01x3+1dx using the Beta function B(a,b)=∫01ta−1(1−t)b−1dtB(a, b) = \int_0^1 t^{a-1}(1-t)^{b-1} dtB(a,b)=∫01ta−1(1−t)b−1dt.
B(1/3,2/3)B(1/3, 2/3)B(1/3,2/3)
23B(1/3,1/2)\frac{2}{3} B(1/3, 1/2)32B(1/3,1/2)
12B(1/2,1/3)\frac{1}{2} B(1/2, 1/3)21B(1/2,1/3)
13B(1/3,1/2)\frac{1}{3} B(1/3, 1/2)31B(1/3,1/2)