According to Wolfram Alpha,
$$\int_0^1 (x^2-1)^{-2/3}\,dx = -3\sqrt{\pi}\frac{\Gamma(1/3)}{\Gamma(-1/6)}$$
Leaving the question of integrability aside, how could I see this? I'm not sure how I would link integrand to integrands of gamma functions - where do I pull of exponential functions from? Sorry, I'm completely new to this kind of integrals so I might not be seeing something obvious
Setting $u=x^2$, we get $$ \int_0^1(x^2-1)^{-\frac{2}{3}}\;dx=\int_0^1(1-x^2)^{-\frac{2}{3}}\;dx=\frac{1}{2}\int_0^1u^{-\frac{1}{2}}(1-u)^{-\frac{2}{3}}\;du=\frac{1}{2}B\Big(\frac{1}{2},\frac{1}{3}\Big) $$ where $B(x,y)$ is the Beta function.
Now using $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, $\Gamma(\frac{1}{2})=\sqrt{\pi}$, and $\Gamma(s+1)=s\Gamma(s)$, it follows that $$ \frac{1}{2}B\Big(\frac{1}{2},\frac{1}{3}\Big)=\frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{1}{3})}{\Gamma(\frac{5}{6})}=-3\sqrt{\pi}\frac{\Gamma(\frac{1}{3})}{\Gamma(-\frac{1}{6})} $$ as claimed.