complex analysis - Why the integral $int_{-infty}^infty dx frac{1}{(x-i)^2} frac{1}{(x+i)^2}$ is not zero?

This might be silly but I am almost pulling my hair.

Why the following integral (along the real axis) is not zero:

$$\int_{-\infty}^\infty dx \frac{1}{(x-i)^2} \frac{1}{(x+i)^2}$$

The integral has no residues and can be completed by a semicircle either in the upper complex plane or the lower complex plane. According to the residue theorem, it should be zero. (Mathematica says it's $\pi/2$)

What's wrong with my logic here?

Edit:
I was somehow given the wrong impression that high-order poles does not contribute to the residue.

Answer

Oh, but the residue theorem does give a nonzero value. There are second-order poles at $\pm i$, and the integrand's residue at such a pole is

$$\lim_{x\to\pm i}\frac{d}{dx}\frac{1}{(x\pm i)^2}=\frac{-2}{(\pm 2i)^3}=\mp\frac{i}{4}.$$ As I think your only mistake was in misunderstanding how higher-order poles provide residues, I'll leave it to you to understand why we can calculate the integral from the pole at $i$ alone, viz. $$\int_{\Bbb R}\frac{dx}{(x^2+1)^2}=2\pi i\times\frac{-i}{4}=\frac{\pi}{2},$$ just as your software said. (It can also be proven with $x=\tan t$.)

Edit: let me motivate the way higher-order poles work. Suppose $f(c)$ is neither $0$ nor infinite. Then

$$\oint\frac{f(z)}{(z-c)^n}dz=\oint\frac{f^{n-1}(c)}{(n-1)!(z-c)}dz=2\pi i\cdot\lim_{z\to c}\frac{f^{n-1}(z)}{(n-1)!}.$$