To compute
$$I =\int \frac{1}{1+x^4} \, dx $$
Put $x = iy$ and we get $I = iI$ from which follows $I = 0$. Compute the integral applying the Cauchy formula to a sectoral contour of radius $R$, one arm of which is along the $x$ axis and the other along the imaginary axis and letting $R \to \infty$. You need to factor $$z^4 +1 = (z−a_1 )(z−a_2 )(z−a_3 )(z−a_4 )$$ and prove that the integral along the circular arc goes to zero.
On
$$ \int\limits_{\big(\text{ray}\big)} \frac{dx} { 1+x^4 } = \int\limits_{\big(\text{ray}/i\big)} \frac{i\,dy}{1+(iy)^2} = i \int\limits_{\big(\text{ray}/i\big)} \frac {dy}{ 1+y^4} $$
As $x$ traverses the ray from $0$ to $\infty$ along the positive real axis, the other variable $y$ goes from $0$ in the direction toward $1/i = -i$ and keeps going beyond that. You can write this in a pleasingly symmetric form that shows that you have a choice of where to divide by $i$: $$ \frac 1 i \int\limits_{\big(\text{ray}\big)} \frac{dx} { 1+x^4 } = \int\limits_{\big(\text{ray}/i\big)} \frac{dx}{1+x^4}. $$
If the posted question regards evaluation of the integral $I$ defined by
$$I=\int_0^\infty \frac{1}{1+x^4}dx$$
then, we can proceed as follows. Let's analyze the contour integral
$$J(R)=\oint_C \frac{1}{1+z^4}dz$$
where $C$ is the contour $C=C_1+C_2+C_3$ and where
$1$ $C_1$ is the line segment from $(0,0)$ to $(R,0)$
$2$ $C_2$ is the circular arc centered at the origin with radius $R$ from $(R,0)$ to $(0,R)$
$3$ $C_3$ is the line segment from $(0,R)$ to $(0,0)$
As $R\to \infty$, the contribution from the integration over $C_2$ vanishes as $O(R^{-3})$.
Thus,
$$\lim_{R\to \infty}J(R) =\int_0^\infty \frac{1}{1+x^4}dx+\int_\infty^0\frac{1}{1+(iy)^4} i \, dy=(1-i)\int_0^\infty\frac{1}{1+x^4}dx$$
Using the residue theorem, we have that (for $R>1$)
$$\begin{align} J(R)&=2\pi i \operatorname{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4}\right)\\\\ &=\lim_{z\to e^{i\pi/4}}\frac{z-e^{i\pi/4}}{1+z^4}\\\\ &=\frac{1}{4e^{i3\pi/4}} \end{align}$$
Putting it all together, we have
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}\frac{1}{1+x^4}dx=\frac{2\pi i}{4(1-i)e^{i3\pi/4}}=\sqrt{2}\pi/4}$$