`int_(-2)^(0){x^(3)+3x^(2)+3x+3+(x+1)cos(x+1)} dx` is equal to

Answer

To solve the integral

\[ \int_{-2}^{0} \left( x^3 + 3x^2 + 3x + 3 + (x+1) \cos(x+1) \right) dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
Notice that the polynomial part \(x^3 + 3x^2 + 3x + 3\) can be rewritten. We can express the polynomial as:

\[ x^3 + 3x^2 + 3x + 3 = (x+1)^3 + 2. \]

This is because:

\[ (x+1)^3 = x^3 + 3x^2 + 3x + 1. \]

So, we can rewrite the integral as:

\[ \int_{-2}^{0} \left( (x+1)^3 + 2 + (x+1) \cos(x+1) \right) dx. \]

### Step 2: Change of variable
Let \(t = x + 1\). Then, \(x = t - 1\) and when \(x = -2\), \(t = -1\), and when \(x = 0\), \(t = 1\). The differential \(dx\) becomes \(dt\). Thus, the integral transforms to:

\[ \int_{-1}^{1} \left( t^3 + 2 + t \cos t \right) dt. \]

### Step 3: Split the integral
Now, we can split the integral into three parts:

\[ \int_{-1}^{1} t^3 dt + \int_{-1}^{1} 2 dt + \int_{-1}^{1} t \cos t dt. \]

### Step 4: Evaluate each part
1. **Evaluate \( \int_{-1}^{1} t^3 dt \)**:
Since \(t^3\) is an odd function, the integral over a symmetric interval around zero is zero:

\[ \int_{-1}^{1} t^3 dt = 0. \]

2. **Evaluate \( \int_{-1}^{1} 2 dt \)**:
This integral is straightforward:

\[ \int_{-1}^{1} 2 dt = 2 \times (1 - (-1)) = 2 \times 2 = 4. \]

3. **Evaluate \( \int_{-1}^{1} t \cos t dt \)**:
Since \(t \cos t\) is also an odd function, the integral over a symmetric interval around zero is zero:

\[ \int_{-1}^{1} t \cos t dt = 0. \]

### Step 5: Combine the results
Now, combining all the results:

\[ \int_{-1}^{1} t^3 dt + \int_{-1}^{1} 2 dt + \int_{-1}^{1} t \cos t dt = 0 + 4 + 0 = 4. \]

Thus, the value of the original integral is

\[ \boxed{4}. \]