A complex line integral

Consider a second curve given by the interval $[-2, 2]$ and combine it with the curve $\mathcal{C}$ . The resulting curve $\mathscr{C}$ will be closed. Therefore, as the only pole within $\mathscr{C}$ is $z=i$ , $I + \int_{-2}^2 \frac{xe^{-x^2}}{x^2+1}\,\mathrm{d}x = \oint_\mathscr{C} \frac{ze^{-z^2}}{z^2+1} \,\mathrm{d} z = 2\pi i \cdot \mathrm{Res}_{z=i} \frac{ze^{-z^2}}{z^2+1}.$ However, note that the function $f: \mathbb{R} \to \mathbb{R}, x \mapsto \frac{xe^{-x^2}}{x^2+1}$ is odd. Therefore, $\int_{-2}^2 \frac{xe^{-x^2}}{x^2+1}\,\mathrm{d}x = 0.$ We conclude that $I =2\pi i \cdot \mathrm{Res}_{z=i} \frac{ze^{-z^2}}{z^2+1} = 2\pi i \cdot \lim_{z\to i} \frac{ze^{-z^2}}{z+i}=\boxed{e\pi i}.$