<s>[INST] How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$? Let $$J_n := \int_{0}^{\infty} \frac{1}{(x^3 + 1)^n} \, {\rm d} x$$
where $n > 2$ is integer. How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$?
 [/INST] Note
\begin{align}
J_{n+1}&= \int_{0}^{\infty} \frac{(x^3+1)-x^3}{(x^3 + 1)^{n+1}} dx
=J_n +\frac1{3n} \int_{0}^{\infty} x\>d\left(\frac{1}{(x^3 + 1)^n} \right)\\
&=J_n - \frac1{3n}J_n =\frac{3n-1}{3n}J_n
\end{align}
</s>