Evaluate for $t\in \mathbb{R}$ $\int_{-\infty}^\infty{e^{itx} \over (1+x^2)^2}dx$

Evaluate for $t\in \mathbb{R}$ $$\int_{-\infty}^\infty{e^{itx} \over (1+x^2)^2}dx.$$

Here is what I have done: Let $f(z)={e^{itz}\over (1+z^2)^2}$. This has two poles $z=i$ $z=-i$ and an essential singularity at $z=- \infty$. When $t\leq 1$ I think the integral becomes ${1\over 2}\pi (1-t)e^t$ and when $t\leq 0$ I think the integral becomes ${1\over 2}\pi(1+|t|)e^{-|t|}$. I'm not sure how to show this. Any solutions or help is greatly appreciated.

Hint: Use contour $\Gamma$ consisting of the upper semicircle $C_\rho^+ : y = \sqrt{\rho^2 - x^2}$ and the line $\gamma$ joining $(-\rho, 0)$ to $(\rho, 0)$. Apply Jordan's lemma to see that the integral vanishes over $C_\rho^+$ (or do this yourself using the M-L estimate).