Answer
$$C\sqrt {2x-1}e^{x}$$
Work Step by Step
$$(e^{x^2} g(x))' = e^{x^2} 2x g(x) + e^{x^2} g'(x) = e^{x^2} 2x g'(x)$$
$$2x g(x) = (2x-1) g'(x)$$
$$\int{g'(x)/g(x)dx} = \int{2x/(2x-1)dx}$$
$$\ln{g(x)} = x+\frac{1}{2} \ln{(2x-1)} +C$$
$$g(x) = C\sqrt {2x-1}e^{x}$$
