1. Apply the chain rule to differentiate the exponential function

$\frac{\text{d}}{\text{d}x}(\text{e}^{x^2+2x}) = (2x+2)\text{e}^{x^2+2x}$
$= 2(x+1)\text{e}^{x^2+2x}$

1. Use the result from part (i) to find the integral of the basic exponential component.

From (i), $\frac{\text{d}}{\text{d}x} (\text{e}^{x^2+2x}) = (2x+2)\text{e}^{x^2+2x} = 2(x+1)\text{e}^{x^2+2x}$

Therefore, integrating both sides gives:
$\int (x+1)\text{e}^{x^2+2x} \text{ d}x = \frac{1}{2}\text{e}^{x^2+2x} + C$

2. Set up integration by parts by appropriately splitting the integrand.

We rewrite the integral to make use of our result from step 1:
$\int (x+1)^3 \text{e}^{x^2+2x} \text{ d}x = \int (x+1)^2 \cdot (x+1)\text{e}^{x^2+2x} \text{ d}x$

Let $u = (x+1)^2$ and $\frac{\text{d}v}{\text{d}x} = (x+1)\text{e}^{x^2+2x}$
Then $\frac{\text{d}u}{\text{d}x} = 2(x+1)$ and $v = \frac{1}{2}\text{e}^{x^2+2x}$

3. Apply the integration by parts formula and simplify the resulting expression.

Using the formula $\int u \text{ d}v = uv - \int v \text{ d}u$:
$\int (x+1)^3 \text{e}^{x^2+2x} \text{ d}x = (x+1)^2 \left( \frac{1}{2}\text{e}^{x^2+2x} \right) - \int \left( \frac{1}{2}\text{e}^{x^2+2x} \right) \cdot 2(x+1) \text{ d}x$
$= \frac{1}{2}(x+1)^2 \text{e}^{x^2+2x} - \int (x+1)\text{e}^{x^2+2x} \text{ d}x$

4. Evaluate the remaining integral and simplify to obtain the final answer.

Since $\int (x+1)\text{e}^{x^2+2x} \text{ d}x = \frac{1}{2}\text{e}^{x^2+2x}$ from our earlier result:
$= \frac{1}{2}(x+1)^2 \text{e}^{x^2+2x} - \frac{1}{2}\text{e}^{x^2+2x} + C$
$= \frac{1}{2}\text{e}^{x^2+2x} \left( (x+1)^2 - 1 \right) + C$
$= \frac{1}{2}\text{e}^{x^2+2x} (x^2+2x+1-1) + C$
$= \frac{1}{2}(x^2+2x)\text{e}^{x^2+2x} + C$