If we complete the square in the exponent, we have
$$\frac 1 2(2x - x^2) = \frac 1 2 \Big(-(x - 1)^2 + 1\Big) = -\left(\frac{x - 1}{\sqrt 2}\right)^2 + \frac 1 2$$
Hence,
\begin{align*}
\int e^{\frac 1 2(2x - x^2)} dx &= \sqrt e \int e^{-\left(\frac{x - 1}{\sqrt 2}\right)^2} dx
\end{align*}
Make the change of variables
$$u = \frac{x - 1}{\sqrt 2},\quad \sqrt 2 \, du = dx$$
and we'll find that our integral is
$$\sqrt{2e} \int e^{-u^2} du$$
Unfortunately, there is no closed form for this integral, but it can then be expressed in terms of the error function, which is defined to be
$$\operatorname{erf}(t) = \frac{2}{\sqrt{\pi}} \int_0^t e^{-u^2} du$$
Alternatively, since you do have bounds on your integral, you should look at the Gaussian integral, which gives that
$$\int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}$$