<s>[INST] prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$ 
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$

My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\frac{5\pi}{8}},e^{i\frac{7\pi}{8}}$ lie in upper semi-circle C and real axis
Given integral value $= 2\pi i \cdot (\text{sum of residues}) = 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right] = 0.27059 \pi$
This is numerically equal to $\frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$. But without using calculator, how to get this expression.
 [/INST] An alternative approach using complex analysis. 
It is not nice to have to calculate many residues. Just reduce it by noting the following:
\begin{align}
I=\int^{\infty}_{-\infty}\frac{x^4}{x^8+1}\,dx=\Re\left(2\int^\infty_{0}\frac{1}{x^4-i}\,dx\right)
\end{align}
Now calculate the RHS by considering:
\begin{align}
\oint_C\frac{1}{z^4-i}\,dz
\end{align}
Where $C$ is a quarter circle  with radius $R>1$ in the first quadrant. There is only one pole inclosed instead of four! Call that number $\alpha$. So you only have to find 1 residue (one fourth of the original number!). By the residue theorem we have:
\begin{align}
\oint_C\frac{1}{z^4-i}\,dz=2\pi i \text{Res}_{z=\alpha}\frac{1}{z^4-i}
\end{align}
On the other hand we have:
\begin{align}
\oint_C\frac{1}{z^4-i}\,dz=-i\int^R_0\frac{1}{x^4-i}\,dx+\int^R_0 \frac{1}{x^4-i}\,dx+\int_{B_R}\frac{1}{z^4-i}\,dz
\end{align}
where $B_R$ is the circular part. After letting $R\to\infty$ you can verify that:
\begin{align}
(1-i)\int^{\infty}_0 \frac{1}{x^4-i}\,dx=2\pi i \text{Res}_{z=\alpha}\frac{1}{z^4-i}
\end{align}
Hence:

\begin{align}
I=\Re\left[4\pi \frac{i}{1-i}\text{Res}_{z=\alpha}\frac{1}{z^4-i} \right]
\end{align}

The calculation are left for you. 
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