<s>[INST] Is it possible to compute $\int_{-\infty}^\infty {x \sin x\over x^4+1}dx$ without using complex analysis? 
$$\int_{-\infty}^\infty {x\sin x\over x^4+1}dx$$

By using things like Feynman's trick (Differentiating under the integral sign) or other high school level calculus concepts? I do not know things like residue theorem.
 [/INST] The same answer as @JackD'Aurizio with slightly more details. I was doing this after he sent his answer so I finish anyway.
Let $\alpha=\frac{1}{2\sqrt[]{2}}$. After partial fraction decomposition we get:
\begin{align}
\int^\infty_{-\infty}\frac{x\sin(x)}{x^4+1}\,dx=-\alpha\int^\infty_{-\infty}\frac{\sin(x)}{x^2+\sqrt[]{2}x+1}\,dx+\alpha\int^\infty_{-\infty}\frac{\sin(x)}{x^2-\sqrt[]{2}x+1}\,dx
\end{align}
Now define:
\begin{align}
I:=\int^\infty_{-\infty}\frac{\sin(x)}{x^2+\sqrt[]{2}x+1}\,dx,\ \ \ \ J:=\int^\infty_{-\infty}\frac{\sin(x)}{x^2-\sqrt[]{2}x+1}\,dx
\end{align}
First we do $I$. Completing the squares in the denominator yields:
\begin{align}
I=\int^\infty_{-\infty}\frac{\sin(x)}{(x+\frac{\sqrt[]{2}}{2})^2+\frac{1}{2}}\,dx
\end{align}
Substitute $u=x+\frac{\sqrt[]{2}}{2}$ and use the formula for $\sin(x+y)$:
\begin{align}
I&=\int^\infty_{-\infty}\frac{\sin\left(u-\frac{\sqrt[]{2}}{2}\right)}{u^2+\frac{1}{2}}\,du\\& = \int^\infty_{-\infty}\frac{\sin\left(u\right)\cos\left(-\frac{\sqrt[]{2}}{2}\right)}{u^2+1}\,du+\int^\infty_{-\infty}\frac{\cos(u)\sin\left(-\frac{\sqrt[]{2}}{2}\right)}{u^2+\frac{1}{2}}\,du\\
 \end{align}
The integral with $\sin(u)$ is convergent and the ingerand is an odd function, so it vanishes. Now we get:
\begin{align}
I=\sin\left(-\frac{\sqrt[]{2}}{2}\right)\int^\infty_{-\infty}\frac{\cos(u)}{u^2+\frac{1}{2}}\,du
\end{align}
This integral can be done with the explicitly mentioned trick by you, namely Feynman's trick. It is done nicely with full details by user Mark Viola in this answer. 
So we get:
\begin{align}
I=\sin\left(-\frac{\sqrt[]{2}}{2}\right) \frac{\pi\sqrt[]{2}}{e^{1/\sqrt[]{2}}}=-\sin\left(\frac{\sqrt[]{2}}{2}\right) \frac{\pi\sqrt[]{2}}{e^{1/\sqrt[]{2}}}
\end{align}
The same can be done for $J$ (do it yourself!). We get:
\begin{align}
J=\sin\left(\frac{\sqrt[]{2}}{2}\right)\frac{\pi\sqrt[]{2}}{e^{1/\sqrt[]{2}}}
\end{align}
Finally we get:

\begin{align}
\int^\infty_{-\infty}\frac{x\sin(x)}{x^4+1}\,dx = \frac{\pi}{e^{1/\sqrt[]{2}}}\sin\left(\frac{\sqrt[]{2}}{2}\right)
\end{align}

</s>