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.