We are going to find the reduction formula by differentiation. $$ \begin{aligned} \frac{d}{d x}\left[\frac{x}{k n\left(x^{k}+1\right)^{n}}\right] &=\frac{1}{k n} \frac{\left(x^{k}+1\right)^{n} \cdot 1-x n\left(x^{k}+1\right)^{n-1} \cdot k x^{k-1}}{\left(x^{k}+1\right)^{2 n}} \\ &=\frac{1}{k n} \cdot \frac{x^{k}+1-k n\left(x^{k}+1-1\right)}{\left(x^{k}+1\right)^{n+1}} \\ &=\frac{1-k n}{k n\left(x^{k}+1\right)^{n}}+\frac{1}{\left(x^{k}+1\right)^{n+1}} \end{aligned}$$
Integrating both sides yields $$ \begin{aligned} \frac{x}{kn\left(x^{k}+1\right)^{n}} &=\frac{1-k n}{k n} \int \frac{d x}{\left(x^{k}+1\right)^{n}}+\int \frac{d x}{\left(x^{k}+1\right)^{n+1}} \\ I_{n+1} &=\frac{x}{kn\left(x^{k}+1\right)^{n}}+\frac{k n-1}{k n} I_{n} . \end{aligned} $$
Furthermore, when we consider the integral with limits $0$ and $\infty$,we got a wonderful closed form for the integral $$ J_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{k}+1\right)^{n}} $$
Applying the reduction formula of $I_n$ with the limits yields $$ \begin{aligned} J_{n+1} &=\frac{k n-1}{k n} J_{n} \\ & \qquad\qquad \vdots \\ &=\frac{(k n-1)[k(n-1)-1] \ldots(k\cdot 1-1)}{k^{n} n !} J_{1} \\ &=\displaystyle \frac{\displaystyle \prod_{j=1}^{n}\left(kj-1\right)}{k^{n} n !} J_{1} \end{aligned} $$
Using the result, $$ J_{1}=\int_{0}^{\infty} \frac{d x}{x^{k}+1}=\frac{\pi}{k} \csc \left(\frac{\pi}{k}\right) $$
Now we can conclude that $$ \boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{k}+1\right)^{n}}=\frac{\pi \displaystyle \prod_{j=1}^{n-1}(k j-1)}{k^{n}(n-1) !} \csc \left(\frac{\pi}{k}\right)} $$
Following your interesting posts for a while, in my humble opinion, it is probably time for you to start using hypergeometric functions or at least the simplest one (the gaussian hypergeometric function).
Using them, you would have $$I_n= \int \frac{d x}{\left(x^{k}+1\right)^{n}}=x \,\, _2F_1\left(\frac{1}{k},n;\frac{k+1}{k};-x^k\right)$$ $$J_n= \int_0^\infty \frac{d x}{\left(x^{k}+1\right)^{n}}=\frac{\Gamma \left(1+\frac{1}{k}\right) \Gamma \left(n-\frac{1}{k}\right)}{\Gamma (n)}$$ which is valid if $\Re(k n)>1\land \Re(k)>0$.
For sure, what you wrote for $J_n$ is the same. With regard to the reduction formula for $I_n$, you will find a lot of them.