How to calculate $\int _{-\infty }^{\infty }\frac{x\sin \left(x\right)}{1+x^4}\,dx$ I want to calculate $\int _{-\infty }^{\infty }\frac{x\sin \left(x\right)}{1+x^4}\,dx$, but I don’t want to use complex analysis.  How can I calculate it?
I tried
$$I\left(t\right)=\int _{-\infty }^{\infty }\frac{x\sin \left(tx\right)}{1+x^4}\,dx$$
$$I''\left(t\right)=-\int _{-\infty }^{\infty }\frac{x^3\sin \left(tx\right)}{1+x^4}\,dx=-\int _{-\infty }^{\infty }\frac{\sin \left(tx\right)}{x}\,dx\:+\int _{-\infty }^{\infty }\frac{\sin \left(tx\right)}{x\left(1+x^4\right)}\,dx$$
$$=-\pi \:+\int _{-\infty }^{\infty }\frac{\sin \left(tx\right)}{x\left(1+x^4\right)}\,dx$$
$$I''''\left(t\right)=-\int _{-\infty }^{\infty }\frac{x\sin \left(tx\right)}{1+x^4}\,dx$$
$$I''''\left(t\right)+I\left(t\right)=0$$
Solving the differential equation and then setting the initial conditions seem like a very long process. How else can I calculate?

With $I\left(t\right)=\int _{-\infty}^{\infty }\frac{x\sin \left(tx\right)}{1+x^4}\:dx$, you have
$I’’’’(t)+I(t)= 0$, along with all the initial conditions
$$I(0)=0, \>\>\>I’(0)=\int_{-\infty}^\infty \frac{x^2}{1+x^4}dx =\frac\pi{\sqrt2} ,\\ I’’(0)=-\pi, \>\>\>
 I’’’(0)=\int_{-\infty}^\infty \frac{1}{1+x^4}dx  =\frac\pi{\sqrt2}\\
$$
which lead to the solution $I(t) =\pi e^{-\frac t{\sqrt2}}\sin\frac t{\sqrt2} $. Thus,
$$\int _{-\infty}^{\infty }\frac{x\sin \left(x\right)}{1+x^4}\:dx
=I(1)=\pi e^{-\frac 1{\sqrt2}}\sin\frac 1{\sqrt2}
$$
