Method #1
Using ##u = x^4##,
\begin{align*}
\int_0^\infty \dfrac{x^2+1}{x^4+1} dx = \frac{1}{4} \int_0^\infty \dfrac{u^{-1/4}+u^{-3/4}}{u+1} du \qquad (*)
\end{align*}
Here:
www.physicsforums.com/threads/exponential-type-integrals.1046843/
in post #12 I proved that, if ##0< a < 1##, then
$$
\int_0^\infty \frac{u^{a-1}}{1+u} du = \frac{\pi}{\sin a \pi}
$$
Using this in ##(*)##, we have
\begin{align*}
\int_0^\infty \dfrac{x^2+1}{x^4+1} dx & = \frac{\pi}{4} \left( \frac{1}{\sin 3 \pi / 4} + \frac{1}{\sin \pi / 4} \right)
\nonumber \\
& = \frac{\pi}{\sqrt{2}}
\end{align*}Method #2:
\begin{align*}
\int_0^\infty \dfrac{x^2 + 1}{x^4 + 1} dx & = \int_0^\infty \dfrac{1 + \dfrac{1}{x^2}}{x^2 + \dfrac{1}{x^2}} dx
\nonumber \\
& = \int_0^\infty \dfrac{1 + \dfrac{1}{x^2}}{\left( x - \dfrac{1}{x} \right)^2 + 2} dx
\nonumber \\
& = \int_0^\infty \dfrac{d \left( x - \dfrac{1}{x} \right)}{\left( x - \dfrac{1}{x} \right)^2 + 2}
\nonumber \\
& = \int_{-\infty}^\infty \dfrac{d u}{u^2 + 2}
\nonumber \\
& = \frac{1}{\sqrt{2}} \int_{-\infty}^\infty \dfrac{d u}{u^2 + 1}
\nonumber \\
& = \frac{\pi}{\sqrt{2}} .
\end{align*}Method #3
\begin{align*}
\int_0^\infty \dfrac{x^2 + 1}{x^4 + 1} d x & = \frac{1}{2} \int_0^\infty \frac{1}{x^2 - \sqrt{2} x + 1} dx + \frac{1}{2} \int_0^\infty \frac{1}{x^2 + \sqrt{2} x + 1} dx
\nonumber \\
& = \frac{1}{2} \int_{-\infty}^\infty \frac{1}{x^2 - \sqrt{2} x + 1} dx
\nonumber \\
& = \frac{1}{2} \int_{-\infty}^\infty \frac{1}{\left( x - \frac{1}{\sqrt{2}} \right)^2 + \frac{1}{2}} dx
\end{align*}
Using ##u = x - \frac{1}{\sqrt{2}}##, the above becomes
\begin{align*}
\int_0^\infty \dfrac{x^2 + 1}{x^4 + 1} d x & = \int_{-\infty}^\infty \frac{1}{2 u^2 + 1} du
\nonumber \\
& = \frac{\pi}{\sqrt{2}} .
\end{align*}