Evaluating $\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} dx$

Question

This question is not a homework question, but I just searched it on WolframAlpha and just want to know how it's done. WolframAlpha lists the following:

$$\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} \mathrm dx = \frac{\sqrt{\pi}\Gamma (\frac{4}{3})}{\Gamma (\frac{5}{6})}$$

How can this be evaluated? I'm not too sure where to start.

Answer 1

Substituting $v=-x$ and $u=v^3$ gives $$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}=\int_0^1\frac{dv}{\sqrt{1-v^3}}=\frac{1}{3}\int_0^1 u^{-\frac{2}{3}}(1-u)^{-\frac{1}{2}}du=\frac{1}{3}\beta\left(\frac{1}{3},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{2}\right)}{3\Gamma\left(\frac{5}{6}\right)} $$ Notice that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma\left(\frac{4}{3}\right)=\frac{1}{3}\Gamma\left(\frac{1}{3}\right)$, we then get $$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}dx=\frac{\sqrt{\pi}\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{5}{6}\right)} $$