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Or

$I(s)=\dfrac{1}{16}\int_0^{\infty}\dfrac{y^{s-\frac34}dy}{1+y}.\tag{1}$

The integral you are looking for is obtained as $I'(0)$ after the change of variables $y=x^4$

Let us make in (1) another change of variables

$\displaystyle t=\frac{y}{1+y}\Longleftrightarrow y=\frac{t}{1-t},dy=\frac{dt}{(1-t)^2}$ This gives

$\begin{aligned} I(s)&=\frac{1}{16}\int_0^1t\cdot\left(\frac{t}{1-t}\right)^{s-\frac74}\cdot \frac{dt}{(1-t)^2}=\\ &=\frac{1}{16}\int_0^1t^{s-\frac34}(1-t)^{-s-\frac{1}{4}}dt=\\& =\frac{1}{16}B\left(s+\frac14,-s+\frac34\right)=\\& =\frac{1}{16}\Gamma\left(s+\frac14\right)\Gamma\left(-s+\frac34\right)=\\ &=\frac{\pi}{16\sin\pi\left(s+\frac14\right)}. \end{aligned}$

Differentiating this with respect to s, we indeed get

$\boxed{I'(0)=-\dfrac{\pi^2\cos\dfrac{\pi}{4}}{16\sin^2\dfrac{\pi}{4}}=-\dfrac{\pi^2\sqrt{2}}{16}}$

Let

$\displaystyle f(n) = \int_0^{\infty} \frac{x^n}{1+x^4} dx$

Where we seek $f'(0)$ .

The integral now is a particular integral of the form :

$\displaystyle \int_0^{\infty} \frac{x^n}{1+x^m} dx \; \; \text{Which evaluates to} \; \; \frac{\pi}{m} \csc (\frac{n+1}{m} \pi )$

Hence:

$\displaystyle f(n) = \frac{\pi}{4} \csc (\frac{n+1}{4} \pi )$

$\displaystyle f'(n) = - \frac{\pi^2}{16} \csc (\frac{n+1}{4} \pi ) \cot (\frac{n+1}{4} \pi ) \overset{n=0}{=} \boxed{-\frac{\pi^2}{8\sqrt{2}} }$

For the sake of completeness, I will prove the particular integral I have used:

$\displaystyle \int_0^{\infty} \frac{x^n}{1+x^m} dx$

$\displaystyle \overset{{\color{#D61F06}{\text{Let} \; u= \frac{1}{1+x^m} }}}{=} \int_1^0 (u)\bigg((\frac{1}{u} -1)^{\frac{1}{m} } \bigg)^n \bigg( -\frac{1}{u^2} \frac{1}{m}(\frac{1}{u} -1)^{\frac{1}{m} -1 } du \bigg)$

$\displaystyle = \frac{1}{m} \int_0^1 u^{-\frac{n+1}{m} } (1-u)^{\frac{n+1}{m} -1 } du \overset{{\color{#D61F06}{\text{Beta Function}}}}{=} \frac{1}{m} B(1-\frac{n+1}{m} , \frac{n+1}{m} )$

$\displaystyle \overset{{\color{#D61F06}{\text{Beta Function Property}}}}{=} \frac{1}{m} \frac{ \Gamma (1-\frac{n+1}{m} ) \Gamma ( \frac{n+1}{m} ) }{\Gamma (1-\frac{n+1}{m} +\frac{n+1}{m} ) }$

$\displaystyle \overset{{\color{#D61F06}{\text{Euler's reflection Formula}}}}{=} \frac{\pi}{m} \csc (\frac{n+1}{m} \pi )$

Simple contour integration application.

$z$ is in the complex plane and the contour is taken as a semicircle of infinte radius above positive real axis.

$\displaystyle \oint{\frac{ln(z)}{1+{z}^{4}}} = \int_{-\infty}^{\infty}{\frac{ln(z)}{1+{z}^{4}}} + \int_{\text{arc}}{\frac{ln(z)}{1+{z}^{4}}}$

The last integral(on RHS) $\rightarrow 0$ as $|z|\rightarrow \infty$ .

Then, the integral we want to find becomes -

$\displaystyle \frac{1}{2} 2\pi\iota\left(\sum{Res\left(\frac{ln(z)}{1+{z}^{4}}, {z}_{i}\right)}\right)$

Evaluating the residues at poles - ${e}^{\iota\pi/4}, {e}^{3\iota\pi/4}$ and summing comes out to be -

$\displaystyle -\frac{{\pi}^{2}}{8\sqrt{2}}$