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Answer

Let

[ f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-t^{2}/2},dt ]

be the standard normal cdf.
For a fixed (x\in\mathbb R) we have to evaluate

[ I(x)=\int_{-\infty}^{\infty}\frac{e^{-t^{2}/2}}{(x-t)^{2}+1},dt . ]

For (a>0)

[ \frac{1}{a^{2}+1}= \int_{0}^{\infty}e^{-a^{2}u},e^{-u},du . ]

Hence

[ I(x)=\int_{0}^{\infty}e^{-u}, \underbrace{\int_{-\infty}^{\infty}e^{-u,t^{2}}e^{ixt},dt}_{=:J(u)},du . ]

The Fourier transform of a Gaussian is again a Gaussian:

[ J(u)=\int_{-\infty}^{\infty}e^{-u,t^{2}}e^{ixt},dt =\sqrt{\frac{\pi}{u}}; e^{-,x^{2}/(4u)} . ]

Thus

[ I(x)=\sqrt{\pi}\int_{0}^{\infty} \frac{e^{-x^{2}/(4u)-u}}{\sqrt{u}};du . ]

Put (u=\dfrac{x^{2}}{4v};(v>0)).
Then (du=-\dfrac{x^{2}}{4v^{2}},dv) and

[ I(x)=\sqrt{\pi},e^{-x^{2}/4}, \frac{x}{2}\int_{0}^{\infty}v^{-3/2}e^{-v},dv . ]

The last integral is a Gamma–function:

[ \int_{0}^{\infty}v^{-3/2}e^{-v},dv =\Gamma!\left(-\tfrac12\right) =-2\sqrt{\pi}. ]

Collecting the constants,

[ I(x)=\sqrt{\pi},e^{-x^{2}/4}, \frac{x}{2},(-2\sqrt{\pi}) =-\pi,x,e^{-x^{2}/4}. ]

[ \boxed{\displaystyle \int_{-\infty}^{\infty}\frac{e^{-t^{2}/2}}{(x-t)^{2}+1},dt =-\pi,x,e^{-x^{2}/4}\qquad (x\in\mathbb R)} ]

The integral is an elementary function; no special functions are required.