Integral of [1+(x^g)]^(-n) from 0 to ∞

\begin{align*}
& \int_0^\infty \dfrac{1}{(1+x^g)^n} dx =
\int_0^\infty \dfrac{(x)^{gn-2}}{(1+x^g)^n} dx \\& \\&\therefore x^g = t \: , \: x = t^{\frac{1}{g}} \Longrightarrow
dx = \dfrac{t^{\frac{1}{g}-1}}{g} dt \\& \\&\therefore \int_0^\infty \dfrac{(x)^{gn-2}}{(1+x^g)^n} dx \Longrightarrow
\int_0^\infty \left[ \dfrac{(t^{\frac{1}{g}})^{gn-2}}{(1+t)^n} \right] \left[ \dfrac{t^{\frac{1}{g}-1}}{g} \right] dt \\& \\&\Longrightarrow \dfrac{\beta \left( n-\frac{1}{g},\frac{1}{g} \right)}{g} = \dfrac{ \left( n - \dfrac{1}{g} \right)! \left( \dfrac{1}{g} \right)!}{ \left( n - \dfrac{1}{g} \right)(n-1)!}\end{align*}

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