Evaluate $\int _{ }^{ }\frac{1}{\sqrt{1+x^3}}dx$

My attempt at solution is...

$ \int _{ }^{ }\frac{1}{\sqrt{1+x^3}}dx\ \left(u=x^3,\ dx=\frac{du}{3x^2}\right) $

$ \int _{ }^{ }\frac{1}{3x^2\sqrt{1+u}}du\ \ \left[\sqrt[3]{u}=x\right] $

$ \int _{ }^{ }\frac{1}{3u^{\frac{2}{3}}\sqrt{1+u}}du $

according to Wolfram this is equal to the function $ _2F_1 $

$ \int _{ }^{ }\frac{1}{3u^{\frac{2}{3}}\sqrt{1+u}}du=\sqrt[3]{u}_2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3}-u\right)+C $

I have no idea how to get to that result, what is the simplest and easiest way to understand to be able to do this type of integrals with the 2F1 function?

Solution 1:

Hint:

$$(1+x^3)^{-1/3}=1-\frac13x^3-\frac13\left(-\frac43\right)\frac{(x^3)^2}2-\frac13\left(-\frac43\right)\left(-\frac73\right)\frac{(x^3)^3}{3!}+\cdots$$ and after integration

$$x\left(1+\frac13\frac{(-x^3)}4+\frac13\frac43\frac{(-x^3)^2}{7\cdot2}+\frac13\frac43\frac73\frac{(-x^3)^3}{10\cdot3!}+\cdots\right)$$