6. Let $\left(x^{2}+2 x-2\right)^{6}=a_{0}+a_{1}(x+2)+a_{2}(x+2)^{2}+\cdots+a_{12}(x+2)^{12}$, where $a_{i}(i=$ $1,2, \cdots, 12)$ are real constants. Then $a_{0}+a_{1}+2 a_{2}+3 a_{3}+\cdots+12 a_{12}=$ $\qquad$ .
6. 64 .

In the given equation, let $x=-2$, we get $a_{0}=64$. Differentiating both sides of the given equation, we get
$$
6\left(x^{2}+2 x-2\right)^{5}(2 x+2)=a_{1}+2 a_{2}(x+2)+3 a_{3}(x+2)^{2}+\cdots+12 a_{12}(x+2)^{11},
$$

Let $x=-1$, substituting in we get $a_{1}+2 a_{2}+3 a_{3}+\cdots+12 a_{12}=0$. Therefore, $a_{0}+a_{1}+2 a_{2}+3 a_{3}+\cdots+12 a_{12}=$ $64+0=64$.