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.

$$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx=2\int_{0}^{\infty} \frac{x^4}{1+x^8} dx=2\int_{0}^{\infty} \frac{(x^8)^{1/2}}{1+x^8} dx\\
=\frac{1}{4} \int_{0}^{\infty} \frac{u^{5/8-1}}{(1+u)^{1}} du=\frac{1}{4}B\left(\frac{5}{8}, 1-\frac{5}{8}\right) =\frac{1}{4}\frac{\Gamma(5/8)\Gamma(1-5/8)}{\Gamma(1) }\\=\color{blue}{\frac{1}{4}\frac{π}{\sin(\frac{5π}{8}) }  =\frac{1}{4}\frac{π}{\sin(\frac{3π}{8}) }}\color{red}{=\frac{π}{\sqrt2}\sin(\frac{π}{8})}  $$ 
where we made use of the beta and Gamma function and the Schwartz duplication formula Along with the the below relations
$$-\frac{\sqrt2}{2}= \cos(\frac{3π}{4}) = \left(\cos^2(\frac{3π}{8})-\sin^2(\frac{3π}{8})\right)=\left(1-2\sin^2(\frac{3π}{8})\right) \\\implies\color{blue}{\sin (\frac{3π}{8}) =\frac{\sqrt{2+\sqrt2}}{2}} $$ 
And by the same token,
 $$\color{red}{\sin (\frac{π}{8}) =\frac{\sqrt{2-\sqrt2}}{2} = \frac{\sqrt{2}}{2\sqrt{2+\sqrt2}} =\frac{\sqrt{2}}{4\sin (\frac{3π}{8}) }.}$$
