`"If " int e^(x^(3)+x^(2)-1)(3x^(4)+2x^(3)+2x)dx=f(x)+C, " then the value of " f(1)xxf(-1) " is"-.`

Explanation

To solve the given integral problem, we start with the expression:

\[ \int e^{(x^3 + x^2 - 1)} (3x^4 + 2x^3 + 2x) \, dx = f(x) + C \]

We need to find the value of \( f(1) \times f(-1) \).

### Step 1: Identify the integral structure
We can rewrite the integral by recognizing that the derivative of \( x^3 + x^2 - 1 \) is \( 3x^2 + 2x \). Thus, we can express the integral as:

\[ \int e^{(x^3 + x^2 - 1)} (3x^4 + 2x^3 + 2x) \, dx = \int e^{(x^3 + x^2 - 1)} \left( x^2 (3x^2 + 2x) + 2x \right) \, dx \]

### Step 2: Split the integral
We can split the integral into two parts:

\[ \int e^{(x^3 + x^2 - 1)} x^2 (3x^2 + 2x) \, dx + 2 \int e^{(x^3 + x^2 - 1)} x \, dx \]

### Step 3: Apply integration by parts
For the first integral, we will use integration by parts. Let:

- \( u = x^2 \)
- \( dv = e^{(x^3 + x^2 - 1)} (3x^2 + 2x) \, dx \)

Then, we compute \( du \) and \( v \):

- \( du = 2x \, dx \)
- \( v = e^{(x^3 + x^2 - 1)} \)

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

### Step 4: Calculate the integral
After applying integration by parts, we find:

\[ f(x) = x^2 e^{(x^3 + x^2 - 1)} - \int e^{(x^3 + x^2 - 1)} (2x) \, dx + 2 \int e^{(x^3 + x^2 - 1)} x \, dx \]

### Step 5: Simplify \( f(x) \)
After simplification, we get:

\[ f(x) = x^2 e^{(x^3 + x^2 - 1)} + C \]

### Step 6: Evaluate \( f(1) \) and \( f(-1) \)
Now we compute \( f(1) \) and \( f(-1) \):

1. **Calculating \( f(1) \)**:
\[ f(1) = 1^2 e^{(1^3 + 1^2 - 1)} = e^{(1 + 1 - 1)} = e^1 = e \]

2. **Calculating \( f(-1) \)**:
\[ f(-1) = (-1)^2 e^{((-1)^3 + (-1)^2 - 1)} = 1 \cdot e^{(-1 + 1 - 1)} = e^{-1} = \frac{1}{e} \]

### Step 7: Calculate \( f(1) \times f(-1) \)
Now, we find:

\[ f(1) \times f(-1) = e \times \frac{1}{e} = 1 \]

### Final Answer
Thus, the value of \( f(1) \times f(-1) \) is:

\[ \boxed{1} \]