Answer
$x=e^{ t(1-t / 2)+c_1}-1$
Work Step by Step
$\begin{aligned} & \frac{d x}{d t}=1-t+x-t x \\ &=1-t+x(1-t) \\ & \frac{d x}{d t}=(1-t)(1+x) \\ & \int \frac{\mathrm{d} x}{1+x}=\int d t(1-t) \\ & \ln |1+x|=t-\frac{t^2}{2}+c \\ & 1+x=e^{ t(1-t / 2)+c_1} \\ & x=e^{t(1-t / 2)+c_1}-1\end{aligned}$
