$$J(c)=\int_{-\infty}^{\infty}\frac{\exp[-(x-c)^2]}{1+x^2}dx=e^{-c^2}\int_{-\infty}^{\infty}\frac{\exp[-x^2]}{1+x^2} e^{2cx}dx$$ The integral on the right can be treated as a the Fourier transform $\mathcal{F}(\exp[-x^2]/(1+x^2))$, with the transform parameter equal to $\mbox{i}2c$. The function is actually symmetric wrt $x$, therefore thus, it is the cosine Fourier transform we are talking about. The necessary transform is available e.g. in volVol. 1 of Bateman & Erdelyi's "Tables of Integral Transforms" (1954). I used a shortcut and computed the transform using Maple. The resulting answer expression is: $$J(c)=\frac{\pi\mbox{e}}{2}\left( \mbox{erfc}(1+\mbox{i}c)e^{\mbox{i}2c}+\mbox{erfc}(1-\mbox{i}c)e^{-\mbox{i}2c} \right)$$ It is easy to check that it this answer satisfies the ODE given obtained by fedja. Whether Written as the sum of conjugate terms, the function $J(c)$ is clearly real-valued for real $c$. It remains an open question whether this is a nicer'' "nicer" form compared to what you had originallyremains an open question!
$$J(c)=\int_{-\infty}^{\infty}\frac{\exp[-(x-c)^2]}{1+x^2}dx=e^{-c^2}\int_{-\infty}^{\infty}\frac{\exp[-x^2]}{1+x^2} e^{2cx}dx$$ The integral on the right can be treated as a Fourier transform $\mathcal{F}(\exp[-x^2]/(1+x^2))$, with the transform parameter equal to $\mbox{i}2c$. The function is actually symmetric wrt $x$, therefore it is the cosine Fourier transform we are talking about. The necessary transform is available e.g. in vol. 1 of Bateman & Erdelyi's "Tables of Integral Transforms" (1954). I used a shortcut and computed the transform using Maple. The resulting answer is: $$J(c)=\frac{\pi\mbox{e}}{2}\left( \mbox{erfc}(1+\mbox{i}c)e^{\mbox{i}2c}+\mbox{erfc}(1-\mbox{i}c)e^{-\mbox{i}2c} \right)$$ It is easy to check that it satisfies the ODE given by fedja. Whether this is a nicer'' form compared to what you had originally remains an open question!