How find this $\int_{1}^{\infty}\frac{1}{x^2\sqrt{x^3-1}}dx$

find The integral $$\int_{1}^{\infty}\dfrac{1}{x^2\sqrt{x^3-1}}dx$$ My try:let $$\sqrt{x^3-1}=t\Longrightarrow x^3=t^2-1$$ Thank I can't,I think this answer maybe use Gamma integral

Sub $x=1/u$, $dx=-du/u^2$. The integral is then equal to

$$\int_0^1 du \left ( \frac{1}{u^3}-1\right )^{-1/2} = \int_0^1 du \, u^{3/2} \, (1-u^3)^{-1/2}$$

Now let $u=v^{1/3}$, $du = \frac13 v^{-2/3} dv$; the integral is then equal to

$$\frac13 \int_0^1 dv \, v^{-2/3} v^{1/2} (1-v)^{-1/2} = \frac13 \int_0^1 dv \, v^{-1/6} (1-v)^{-1/2}$$

This you may recognize as a Beta function:

$$\frac13 \frac{\Gamma(5/6) \Gamma(1/2)}{\Gamma(4/3)} = \frac{\sqrt{\pi} \, \Gamma(5/6)}{\Gamma(1/3)}$$

Note: one may further simplify this by using the duplication and reflection formulae to express this integral in terms of $\Gamma(1/3)^3$, which reduces to a value of the complete elliptic integral of the first kind.

With your substitution $\sqrt{x^3-1}=t$ the integral becomes $$\int_1^{\infty} \frac{1}{x^2 \sqrt{x^3-1}} dx =\frac{1}{3} \int_0^{\infty} \frac{t^{-1/2}}{(1+t)^{4/3}} dt=\frac{1}{3}B\left(\frac{1}{2},\frac{5}{6}\right)$$ taking into account that the beta function is $B\left(p,q\right)=\int_0^{\infty} \frac{t^{p-1}}{(1+t)^{p+q}} dt$ (see http://dlmf.nist.gov/5.12#E3).