Putting $ \displaystyle=4, \alpha=2, n=2 $ in my post,
$$\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}dx=\frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)$$
we can conclude that $$ \begin{aligned} I &=\frac{\pi}{4(2-1) !} \csc \frac{3 \pi}{4}\left(1-\frac{3}{4}\right) \\ &=\frac{\sqrt{2} \pi}{16} \end{aligned} $$
Question:
Is there any other method?
Your suggestion and alternative are warmly welcome.
Integrate both sides of the equation $\left(\frac{x^3}{1+x^4}\right)’= \frac{4x^2}{(1+x^4)^2}-\frac{x^2}{1+x^4}$ to obtain $$\int_{0}^{\infty} \frac{x^{2}}{(1+x^{4})^{2}} d x = \frac14\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}} d x=\frac14 \cdot \frac\pi{2\sqrt2}=\frac\pi{8\sqrt2} $$
Letting $x\mapsto \frac{1}{x} $ yields $$ I=\int_{0}^{\infty} \frac{x^{4}}{\left(x^{4}+1\right)^{2}} d x $$
Adding them together gives $$ \begin{aligned} 2 I &=\int_{0}^{\infty} \frac{x^{2}+x^{4}}{\left(x^{4}+1\right)^{2}} d x \\ &=\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}\right)^{2}} d x \\ &=\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^{2}+2\right]^{2}} \end{aligned} $$ Letting $x-\frac{1}{x}=\sqrt{2} \tan \theta$, we have
$$ I=\frac{\sqrt{2}}{8} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} \theta d \theta= \frac{\sqrt{2} \pi}{16} $$ We can conclude that
You can use residue theorem. Start by extending the integral from $-\infty$ to $+\infty$, which you can do since the function is even.
Then, use the semi-circle in the complex plane for integration: $\Gamma(\theta) = Re^{i\theta}$, with $\theta \in [0,\pi]$. You will get the desired result in the limit $R\rightarrow +\infty$.
In that limit, the integral in the semi-circle vanishes due to Jordan Lemma. The residues inside the contour are $e^{i\pi/4}$ and $e^{i3\pi/4}$, with Res$\big(\frac{z^2}{(z^4+1)^2},z=e^{i\pi/4}\big)=\frac{e^{3i\pi/4}}{16}$ and Res$\big(\frac{z^2}{(z^4+1)^2},z=e^{i3\pi/4}\big)=\frac{e^{5i\pi/4}}{16}$. The result is then
$\int_{0}^{+\infty}dx \frac{x^2}{(x^4+1)^2} =\frac{1}{2} \int_{-\infty}^{+\infty}dx \frac{x^2}{(x^4+1)^2}=\pi i \sum \text{Res} = \frac{\pi i}{8}(e^{5i\pi/4} + e^{3i\pi/4} ) = \frac{\pi}{8\sqrt{2}} = \frac{\sqrt{2}\pi}{16} $
Consider $\mathcal I(a)=\displaystyle\int_0^\infty\frac{x^2}{x^4+a^4}\mathrm dx, a>0$. Then, using Glasser's master theorem,
$$\mathcal I(a)=\int_0^\infty\frac{\mathrm dx}{\left(x-\frac{a^2}x\right)^2+2a^2}=\int_0^\infty\frac{\mathrm dx}{x^2+2a^2}=\frac{\pi}{2\sqrt2a}$$ $$\implies\int_0^\infty\frac{x^2}{(x^4+a^4)^2}\mathrm dx=\frac{\pi}{8\sqrt2a^5}$$
$$I=\int_{0}^{\infty} \frac{x^{2}}{\left(1+x^{4}\right)^{2}} d x=\int_{0}^{\infty} \frac{x^{2}}{x^8+2x^4+1} d x$$
Using the below result, $${\color{red}{F(s,\alpha)=\int_0^\infty \frac{x^{s-1}}{x^2+2x\cos\alpha +1}\,dx=\frac{\pi\sin((1-s)\alpha)}{\sin\alpha\sin(\pi s)}}}$$
$$\int_0^\infty \frac{x^{s-1}}{x^2 + 2x\cos\alpha + 1}\,dx \underset{x\to x^4}= 4\int_0^\infty \frac{x^{4s - 1}}{x^8 + 2x^4\cos\alpha + 1} \, dt$$
$$\implies{F(s,\alpha)=\int_0^\infty \frac{x^{4s - 1}}{x^8 + 2x^4\cos\alpha + 1} \, dt=\frac{\pi\sin((1-s)\alpha)}{4\sin\alpha\sin(\pi s)}}$$
$${\therefore{I=\lim_{\alpha\to 0}\lim_{s\to 3/4}F\left(s,\alpha\right)=\lim_{\alpha\to 0}\lim_{s\to 3/4}\frac{\pi\sin((1-s)\alpha)}{4\sin\alpha\sin(\pi s)}=\frac{\pi}{8\sqrt2}}}$$
Using $$ \frac1{(1+x)^2}=\sum_{n=0}^\infty(n+1)(-x)^n=\sum_{n=0}^\infty\frac{1}{n!}\varphi(n)(-x)^n $$ with $$ \varphi(x)=\Gamma(x+2). $$ one has $$ \int_0^\infty\frac{x^4}{(1+x^4)^2}dx\overset{x^4\to x}=\frac14\int_0^\infty x^{\frac14}\frac{1}{(1+x)^2}dx\\ =\frac14\int_0^\infty x^{\frac54-1}\frac{1}{(1+x)^2}dx\\ =\frac14\Gamma(\frac54)\Gamma(-\frac54+2)=\frac1{4}\Gamma(\frac54)\Gamma(\frac34)=\frac1{16}\cdot\frac{\pi}{\sin(\frac\pi4)}=\frac{\pi\sqrt2}{16} $$ by Ramarujan's Master Theorem (https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem).
