Contour integral is calculated correctly, but then with integral
$$\int_{-\infty}^\infty \frac{xe^{ix}}{1+x^4}dx$$
you are wrong: $e^{-ix}$ appears only after you transform it
$$\int_{-\infty}^\infty \frac{xe^{ix}}{1+x^4}dx= \int_0^\infty \frac{xe^{ix}}{1+x^4}dx +\int_{-\infty}^0 \frac{xe^{ix}}{1+x^4}dx = \int_0^\infty \frac{xe^{ix}}{1+x^4}dx +\int_{\infty}^0 \frac{-xe^{-ix}}{1+x^4}dx = \int_0^\infty \frac{x(e^{ix}-e^{-ix})}{1+x^4}dx = 2i I$$ .