<s>[INST] Compute integral $\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$ Compute the integral. 
$$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$
The answer at the back of the book is 
$$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
 [/INST] Note
\begin{align}
\int_{-\infty}^\infty \frac{x^4}{1+x^8} dx
& =-\frac1{2i}\int_{-\infty}^\infty \left(\frac{1}{1+i x^4} - \frac{1}{1-i x^4} \right)dx \\
& = -\text{Im} \int_{-\infty}^\infty \frac{1}{1+i x^4}dx
\overset{t=i^{1/4}x} =  -\text{Im} \left( e^{-i\frac\pi8}\int_{-\infty}^\infty \frac{1}{1+t^4}dt\right)\\
&=   -\text{Im} \left( e^{-i\frac\pi8}\cdot \frac\pi{\sqrt2} \right)= \frac\pi{\sqrt2}\cdot\sin\frac\pi8= \frac\pi2\sqrt{1-\frac1{\sqrt2}}
\end{align}
</s>