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

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Calculate the integral $$ \int_{-6}^{-2} (x+4)e^{x^2+8x+12}dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{-6}^{-2} \left(x+4ight)e^{x^{2}+8x+12} dx$$
- step1: Evaluate the integral:
  $$\int \left(x+4ight)e^{x^{2}+8x+12} dx$$
- step2: Use the substitution $$dx=\frac{1}{2x+8} dt$$ to transform the integral$$:$$
  $$\int \left(x+4ight)e^{x^{2}+8x+12}	imes \frac{1}{2x+8} dt$$
- step3: Simplify:
  $$\int \frac{e^{x^{2}+8x+12}}{2} dt$$
- step4: Use the substitution $$t=x^{2}+8x+12$$ to transform the integral$$:$$
  $$\int \frac{e^{t}}{2} dt$$
- step5: Rewrite the expression:
  $$\int \frac{1}{2}e^{t} dt$$
- step6: Use properties of integrals:
  $$\frac{1}{2}	imes \int e^{t} dt$$
- step7: Evaluate the integral:
  $$\frac{1}{2}e^{t}$$
- step8: Multiply the terms:
  $$\frac{e^{t}}{2}$$
- step9: Substitute back:
  $$\frac{e^{x^{2}+8x+12}}{2}$$
- step10: Return the limits:
  $$\left(\frac{e^{x^{2}+8x+12}}{2}ight)\bigg |_{-6}^{-2}$$
- step11: Calculate the value:
  $$0$$
The integral $$ \int _ { - 6 } ^ { - 2 } ( x + 4 ) e ^ { x ^ { 2 } + 8 x + 12 } d x $$ evaluates to 0.
The integral $$ \int_{-6}^{-2} (x+4)e^{x^2+8x+12}dx $$ equals 0.

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