Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and $a$ is positive.
First of all, let’s start with the ‘simple’ case, $$
I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}.
$$
However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula
$$
\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
which I can’t prove by elementary method yet. Please tell me if you have any.
Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]} 
=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$
Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right)
\cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}}
\end{aligned}
$$
Rearranging and simplifying yields
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi  \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)}
$$
Putting $a=1$ gives our formula
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi  \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)}
$$
For verification, $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26 
=\frac{1118260 \pi}{4782969 \sqrt{3}},
\end{aligned}
$$
which is checked by Wolframalpha
$$\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\
&=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\
&\doteq 0.845906950943631,
\end{aligned}
$$
which is checked by Wolframalpha
My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$

Glad to see that there are several nice solutions to the problem, now I want to add one more by discovering a reduction formula using integration by parts.
In order to use integration by parts, I applied the inversion substitution to the original integral.
For any fixed natural number $n$, we now define
$$
J_{m}:=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} \stackrel{x \rightarrow \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{m n-2}}{\left(x^{n}+1\right)^{m}} d x
$$
Noting that $$
\frac{d}{d x}\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)=-(m-1) \cdot \frac{nx^{n-1}}{\left(x^{n}+1\right)^{m}},
$$
we have
$$J_m=-\frac{1}{n(m-1)} \int_{0}^{\infty} x^{m n-n-1} d\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)
$$
By integration by parts, we get
$$
\begin{aligned}
J_{m} &\left.=-\frac{m-1}{n}\left[\frac{x^{m n-n-1}}{\left(x^{n}+1\right)^{m-1}}\right]_{0}^{\infty}+\frac{1}{n(m-1)} \int_{0}^{\infty} \frac{(m n-n-1) x^{m n-1-2}}{\left(x^{n}+1\right)^{m-1}}\right] \\
&=\frac{m n-n-1}{n(m-1)} \int_{0}^{\infty} \frac{x^{m n-n+2}}{\left(x^{n}+1\right)^{m-1}} d x\\&= \frac{\left(m-1-\frac{1}{n}\right)}{m-1} J_{m-1}
\end{aligned}
$$
Applying the reduction formula $m$ times yields inductively
\begin{aligned}
J_m&=\frac{m-1-\frac{1}{n}}{m-1} \cdot \frac{m-2-\frac{1}{n}}{m-2} \cdot \frac{m-3-\frac{1}{n}}{m-3} \cdot \frac{1-\frac{1}{n}}{1} J_{0} \\
& = \frac{1}{(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) J_{0}
\end{aligned}
Using the well-known formula,
$$
J_0=\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),
$$
we can conclude that
$$\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} =\frac{\pi \csc \left(\frac{\pi}{n}\right)}{n(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) $$
