After investigating the indefinite integral in the post, I found that
$$\int \frac { d x } { x ^ { 4 } + 1 } = \frac { 1 } { 4 \sqrt { 2 } } \left[ 2 \tan ^ { – 1 } \left( \frac { x ^ { 2 } – 1 } { \sqrt { 2 } x } \right) + \ln \left| \frac { x ^ { 2 } + \sqrt { 2 } x + 1 } { x ^ { 2 } – \sqrt { 2 } x + 1 } \right|\right] + C;$$
$$\int \frac{d x}{\left(1+x^{4}\right)^{2}}=\frac{x}{4\left(1+x^{4}\right)}+\frac{3}{16 \sqrt{2}}\left[2 \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2} x}\right)+\ln \left|\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right|\right]+C
$$
Plugging limits yields
$$I_1= \int_0^{\infty} \frac{d x}{1+x^4}=\frac \pi{2\sqrt2};\quad
I_2=\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^2}=\frac{3 \pi}{8 \sqrt{2}}
$$
I guess about the value of the general integral
$$I_n=\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^n} =\frac {p\pi}{q\sqrt 2} $$
The form of the integrand reminds me that Beta function may help. Then I put the substitution $y=x^4$ and get
$$
I_n =\frac{1}{4} \int_0^{\infty} \frac{y^{-\frac{3}{4}} d y}{(1+y)^n} =\frac{1}{4} B\left(\frac{1}{4}, n-\frac{1}{4}\right)
$$
Expressing in terms of Gamma function gives
$$
\begin{aligned}I_n&=\frac{1}{4} \frac{\Gamma\left(\frac{1}{4}\right) \Gamma\left(n-\frac{1}{4}\right)}{\Gamma(n)} \\&= \frac{1}{4(n-1) !} \Gamma\left(\frac{1}{4}\right) \Gamma\left(n-\frac{1}{4}\right)\\
&=\frac{1}{4(n-1)!} \Gamma\left(\frac{1}{4}\right)\left(n-1-\frac{1}{4}\right)\left(n-2-\frac{1}{4}\right) \cdots \frac{3}{4} \Gamma\left(\frac{3}{4}\right) \cdots (*) \\
&=\frac{1}{4(n-1) !} \left[\frac{1}{4^{n-1}}(4 n-5)(4 n-9) \cdots 3 \right] \pi \csc \left(\frac{\pi}{4}\right) \cdots (**)\\
&=\frac{\pi \sqrt{2}}{4^n(n-1) !} \prod_{k=1}^{n-1}(4 k-1)
\end{aligned}
$$
(*) uses $\Gamma(z+1)=z \Gamma(z)$ for all $z\in \mathbb{C}$ and
(**) uses $\Gamma(x) \Gamma(1-x)=\pi \csc (\pi x)$ for $z\not \in \mathbb{Z}.$
For examples,
$$
\begin{aligned}
& I_3=\frac{\pi \sqrt{2}}{128} \cdot 3 \cdot 7=\frac{21 \pi \sqrt{2}}{128} \\
& I_4=\frac{\pi \sqrt{2}}{256 \cdot 6} \cdot 3 \cdot 7 \cdot 11=\frac{77 \pi \sqrt{2}}{512}\\& I_{10}=\frac{\pi \sqrt{2}}{4^{10} \cdot 9 !} \prod_{k=1}^{9}(4 k-1)=\frac{15646785 \pi \sqrt{2}}{134217728}
\end{aligned}
$$
checked by WA.
My Question: Are there any alternative methods?
Your comments and alternative solutions are highly appreciated.
Best Answer
Any standard method (including scaling $x$ in the first display equation in the question statement) yields $$\int_0^\infty \frac{dx}{a + x^4} = \frac{\pi}{2 \sqrt 2} a^{-3 / 4} .$$ If $n$ is a positive integer, differentiating gives \begin{align} \frac{d^{n - 1}}{da^{n - 1}} \int_0^\infty \frac{dx}{a + x^4} &= \frac{d^{n - 1}}{da^{n - 1}} \left(\frac{\pi}{2 \sqrt 2} a^{-3 / 4} \right) \\ (-1)^n (n - 1)!\int_0^\infty \frac{dx}{(a + x^4)^n} &= \frac{\pi}{2 \sqrt 2} \underbrace{\left(-\frac{3}{4}\right)\left(-\frac{7}{4}\right) \cdots \left(\frac54 - n \right)}_{n - 1} a^{\frac{1}{4} - n} . \end{align} Evaluating at $a = 1$ and rearranging gives $$\int_0^\infty \frac{dx}{(1 + x^4)^n} = \frac{\pi \sqrt 2}{4^n (n - 1)!} \prod_{k = 1}^{n - 1} (4 k - 1),$$ which agrees with your solution.