Suppose ${{\bf{\xi }}\sim N({\bf{0}},{\bf{I}})}$, then $${\bf{x}} = {\bf{\hat x}} + {{\bf{P}}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}{\bf{\xi }}$$ where ${\bf{P}}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. 2}}}$ is a lower triangle matrix or the Cholesky decomposition of $\bf{P}$ such that ${{\bf{P}}^{\frac{1}{2}}}{({{\bf{P}}^{\frac{1}{2}}})^T} = {\bf{P}}$.
By changing the variable of integration from $x$ to $\xi$ we must get (1)
$$\int_{{{\bf{x}}_1}}^{{{\bf{x}}_2}} {{\bf{g}}({\bf{x}})N({\bf{x}};{\bf{\hat x}},{\bf{P}})} d{\bf{x}} = \int_{_{{\bf{\xi }}1}}^{{{\bf{\xi }}_2}} {{\bf{g}}({\bf{\hat x}} + {{\bf{P}}^{\frac{1}{2}}}{\bf{\xi }})N({\bf{\xi }};{\bf{0}},{\bf{I}})} d{\bf{\xi }}$$.
However I think the correct substutaion is $$\begin{array}{l} {\bf{x}} = {\bf{\hat x}} + {{\bf{P}}^{\frac{1}{2}}}{\bf{\xi }} \to d{\bf{x}} = d({{\bf{P}}^{\frac{1}{2}}}{\bf{\xi }}) = d{\bf{\xi }}{{\bf{P}}^{\frac{1}{2}}} = {({{\bf{P}}^{\frac{1}{2}}})^T}d{\bf{\xi }}\\ \int_{{{\bf{x}}_1}}^{{{\bf{x}}_2}} {{\bf{g}}({\bf{x}})N({\bf{x}};{\bf{\hat x}},{\bf{P}})} d{\bf{x}}\mathop = \limits^{{\bf{x}} = {\bf{\hat x}} + {{\bf{P}}^{\frac{1}{2}}}{\bf{\xi }}} \int_{_{{\bf{\xi }}1}}^{{{\bf{\xi }}_2}} {{\bf{g}}({\bf{\hat x}} + {{\bf{P}}^{\frac{1}{2}}}{\bf{\xi }})N({\bf{\xi }};{\bf{0}},{\bf{I}})} \underbrace {{{({{\bf{P}}^{\frac{1}{2}}})}^T}}_{{\rm{???}}}d{\bf{\xi }} \end{array}$$ So, what will happen to the term ${{{({{\bf{P}}^{\frac{1}{2}}})}^T}}$?
Since we are doing a change of variables in more than 1 dimension, it is true that
$$d\mathbf{x}=J(\mathbf{x}, \mathbf{\xi})d\mathbf{\xi}$$
where $J$ is the jacobian determinant of the transformation:
$$J=|\det(\frac{\partial x_i}{\partial \xi_j})|$$
In your case
$$\frac{\partial x_i}{\partial \xi_j}=P^{1/2}_{ij}$$
and thus $J=|\det(P^{1/2})|=\sqrt{\det(P)}$, which in turn cancels with the normalization of the multivariate Gaussian.
EDIT:
The multivariate normal distribution is given by the formula:
$$N(\mathbf{x},\mathbf{y},\mathbf{P})=\frac{1}{\sqrt{(2\pi)^n \det\mathbf{P}}}\exp(-(\mathbf{x-y})^T\mathbf{P}^{-1}(\mathbf{x-y}))$$
The Jacobian cancels the denominator to produce the formula for $N(\mathbf{x,0,I})=\frac{1}{(2\pi)^{n/2}}e^{-|\mathbf{x-y}|^2}$ after the change of variables.