Stein's identity with Normal random variable I'm having trouble to understand the Stein's identity in Theory of point estimation

Let $X$ a random variable with density in canonical form given by 
  $$p(x|\eta)=\exp\Big(\sum \eta_iT_i(x)-A(\eta)\Big)h(x)$$ where $\eta$ 
  is the natural parameter, and $T(x)$ is some statistic. Take some
  differentiable function $g$ such that $\mathbb{E}[g'(X)]<\infty$, then
$$\mathbb{E}\left(\left[\frac{h'(X)}{h(X)}+\sum \eta_iT_i'(X)\right]g(X)\right)=-\mathbb{E}[g'(X)] \tag{1}$$

Example: If $X\sim N(\mu,\sigma^2)$ then the above expression becomes
$$\mathbb{E}[g(X)(X-\mu)]=\sigma^2\mathbb{E}[g'(X)] \tag{2}$$
where $\mathbb{E}(X)=\mu$ if $g(x)=1$ and $\mathbb{E}(X^2)=\sigma^2+\mu^2$
Anyone can help me figure out how they reach equation (2) from equation (1)?
I do not know if I'm letting anything go, but what's the idea behind this lemma? Why would I want to calculate the mean of the expression on the left side in equation (1)?
 A: For your first question,
for the normal density with mean $\mu$ and variance $\sigma^2$ we have, in order to map it to the  Exponential Family form,
$$\eta = \frac {\mu}{\sigma}$$
$$ T(x) = \frac {x}{\sigma} \implies T'(x) = \frac {1}{\sigma}$$
$$h(x) = \frac {1}{\sigma}\phi(x/\sigma) \implies h'(x) = h(x) \cdot \left(-\frac{x}{\sigma^2}\right)$$
where $\phi()$ is the standard normal density.
A: The result is mostly meaningful in the multidimensional case:
$$\mathbb{E}[||X-g(X)X-\mu||^2]=\mathbb{E}[||X-\mu||^2+||g(X)X||^2-2g(X)X^{\text{T}}(X-\mu)]$$implies terms of the form
$$\int_{\mathbb{R}^p} g(x)x_i(x_i-\mu_i) \varphi(x-\mu)\,\text{d}x$$
where $\varphi$ is the joint density. An inner integral in this multivariate integral is
$$\int_{-\infty}^{+\infty}g(x)x_i(x_i-\mu_i) \frac{\exp\{-(x_i-\mu_i)^2/2\sigma^2\}}{\sqrt{2\pi}\sigma}\,\text{d}x_i=\int_{-\infty}^{+\infty}g(x)x_i\sigma^2\dfrac{\text{d}}{\text{d}x_i}\frac{-\exp\{-(x_i-\mu_i)^2/2\sigma^2\}}{\sqrt{2\pi}\sigma}\,\text{d}x_i=\int_{-\infty}^{+\infty}\sigma^2\frac{\exp\{-(x_i-\mu_i)^2/2\sigma^2\}}{\sqrt{2\pi}\sigma}\frac{\partial}{\partial x_i}g(x)x_i\,\text{d}x_i=\int_{-\infty}^{+\infty}\sigma^2\frac{\exp\{-(x_i-\mu_i)^2/2\sigma^2\}}{\sqrt{2\pi}\sigma}\left\{g(x)+x_i\frac{\partial}{\partial x_i}g(x)\right\}\,\text{d}x_i$$
Therefore
$$\mathbb{E}[g(X)X^{\text{T}}(X-\mu)]=\sum_{i=1}^p\mathbb{E}[g(X)+X_i\frac{\partial}{\partial x_i}g(X)]=\mathbb{E}[pg(X)+X^{\text{T}}\nabla g(X)]$$
This implies that$$\mathbb{E}[||X-g(X)X-\mu||^2]=\mathbb{E}[||X-\mu||^2]+\mathbb{E}[||g(X)X||^2]-2\sigma^2\mathbb{E}[pg(X)+X^{\text{T}}\nabla g(X)]$$
Therefore, if 
$$||g(x)x||^2-2\sigma^2pg(x)-\sigma^2x^{\text{T}}\nabla g(x)\le 0$$
for all $x$'s the estimator $X-g(X)X$ does better than $X$ for all $\mu$'s
