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 $J_{n+1} = \frac{3n-1}{3n} J_n$?
 A: 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 $\dfrac x{(x^3+1)^m}\big|_0^\infty=0-0$
A: 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}
A: Beware: overkill. By letting $\frac{1}{1+x^3}=u$ and exploiting Euler's Beta function we have
$$ J_n = \frac{1}{3}\int_{0}^{1} u^{n-4/3}(1-u)^{-2/3}\,du = \frac{1}{3}\,B\left(n-\frac{1}{3},\frac{1}{3}\right)=\frac{\Gamma\left(n-\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)}{3\,\Gamma(n)}$$
hence
$$ \frac{J_{n+1}}{J_n} = \frac{n-\frac{1}{3}}{n}=\frac{3n-1}{3n}=1-\frac{1}{3n}.$$

Since $J_1=\frac{2\pi}{3\sqrt{3}}$ by partial fraction decomposition we also have
$$ J_n=\frac{2\pi}{3\sqrt{3}}\prod_{k=1}^{n-1}\left(1-\frac{1}{3k}\right) $$
for any $n\geq 2$. $J_n$ is a log-convex function by the Cauchy-Schwarz inequality and $J_n$ behaves like $\frac{\Gamma(1/3)}{3\sqrt[3]{n}}$ for large values of $n$.
