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?

The differential equation by itself is not bad
$$I(t)=e^{-\frac{t}{\sqrt{2}}} \left(\left(c_1 e^{\sqrt{2} t}+c_2\right) \sin
   \left(\frac{t}{\sqrt{2}}\right)+\left(c_3 e^{\sqrt{2} t}+c_4\right) \cos
   \left(\frac{t}{\sqrt{2}}\right)\right)$$ but, as you wrote, the problem could be to set the conditions (but you can do it)
Using algebra, let $a,b,c,d$ be the roots of $x^4+1=0$ (you know them). So
$$\frac x{x^4 +1}=\frac x{(x-a)(x-b)(x-c)(x-d)}$$ Use patial fraction decomposition and face four integrals looking like
$$I_k=\int_{-\infty}^\infty\frac {\sin(x)}{x-k} dx\qquad \text{where} \qquad \text{k   is a complex number}$$ Make $x=t +k$
$$\frac {\sin(x)}{x-k}=\frac {\sin(t+k)}{t}=\cos(k)\frac {\sin(t)}{t}+\sin(k)\frac {\cos(t)}{t}$$ and we shall face sine and cosine integrals. But the end result is simple
$$I_k=\pi \, e^{i k}$$
