Evaluating a Real Improper Integral by Residues I am having trouble evaluating this improper integral due to its integrand and the singularities that are present. The question reads as > Show that $\int_{-\infty}^{\infty}\frac{dx}{x^4+1}=\frac{\pi}{\sqrt{2}}$. The contour is assumed to be the boundary of the half disc $|z|\leq R, Im(z)\geq 0$ taken once anticlockwise. You may assume that the integrals converge. The questions that I have been solving have had "nice" singularities; in the form of $x+iy,x,y\in \mathbb{R}$. I understand the process that takes place in this case. However, this particular integrand has singularities (inside the boundary) of $z=i^{1/2},i^{3/2}$. This is already an intuitive problem. How does one situate these singularities inside the boundary and thus progress with calculations? Furthermore, if I continue with the methodic process I will not get the answer. Thank you in advanced for you help.

Factorization is $(x+0.5+0.5i) \sqrt2)(x-0.5+0.5i) \sqrt2)(x+0.5-0.5i)\sqrt2)(x-0.5-0.5i) \sqrt2)$ The third and fourth factor provide the root in the upper half plane. To find the residues: Knock out the third factor and substitute its roots in the fraction. That's one residue. And then, knock out the fourth factor (put the third back of course) and then put in its root in the fraction. Now you have two residues. Add them up and multiply by $2 \pi i$. That should do it. It's time for me to sleep now. Good luck