<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] Hint:
Let $y=\dfrac x{(x^3+1)^m}$
$$\dfrac{dy}{dx} =\dfrac1{(x^3+1)^m}+\dfrac{(-m)x(3x^2)}{(x^3+1)^{m+1}} =\cdots =\dfrac{3m}{(x^3+1)^{m+1}}-\dfrac{3m-1}{(x^3+1)^m}$$
Integrate both sides wrt $$\dfrac x{(x^3+1)^m}=3mI_{m+1}-(3m-1)I_m$$ where $$I_n=\int\dfrac{dx}{(x^3+1)^n}$$
Now $\dfrac x{(x^3+1)^m}\big|_0^\infty=0-0$
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