\( \int _ { - 6 } ^ { - 2 } ( x + 4 ) e ^ { x ^ { 2 } + 8 x + 12 } d x = \square \)

To solve the integral \( \int _ { - 6 } ^ { - 2 } ( x + 4 ) e ^ { x ^ { 2 } + 8 x + 12 } d x \), we start by simplifying the expression inside the integral. Notice that \( x^2 + 8x + 12 = (x + 4)^2 \). Then, we can rewrite the integral: \[ \int_{-6}^{-2} (x + 4) e^{(x + 4)^2} \, dx \] Next, we perform a substitution to simplify further. Let \( u = x + 4 \), then \( du = dx \), and changing the limits gives \( u = -2 \) when \( x = -6 \) and \( u = 2 \) when \( x = -2 \). Now the integral becomes: \[ \int_{-2}^{2} u e^{u^2} \, du \] This integral can be solved using integration by parts where we let \( v = e^{u^2} \) and \( dv = 2u \, du \). The integral evaluates to: \[ \left[ \frac{1}{2} e^{u^2} \right]_{-2}^{2} \] Calculating this gives: \[ \frac{1}{2} (e^{4} - e^{4}) = 0 \] Thus, the value of the integral is \( 0 \).