If `f(x)=int_(-2)^(x)(t^(4)-bt^(3)+(b+1)t^(2)-bt+b)dt` strictly increases `AAxepsilonR` then no. of integers in range of `b` is

Explanation

To solve the problem, we need to analyze the function \( f(x) = \int_{-2}^{x} (t^4 - bt^3 + (b+1)t^2 - bt + b) \, dt \) and determine the conditions under which this function is strictly increasing.

### Step-by-Step Solution:

1. **Understanding Strictly Increasing Function**:
A function \( f(x) \) is strictly increasing if its derivative \( f'(x) > 0 \) for all \( x \in \mathbb{R} \).

2. **Using the Leibniz Rule**:
To find \( f'(x) \), we apply the Leibniz rule for differentiation under the integral sign:
\[ f'(x) = \frac{d}{dx} \int_{-2}^{x} (t^4 - bt^3 + (b+1)t^2 - bt + b) \, dt = (x^4 - bx^3 + (b+1)x^2 - bx + b) \]

3. **Setting the Derivative Greater Than Zero**:
We need to ensure that:
\[ x^4 - bx^3 + (b+1)x^2 - bx + b > 0 \quad \forall x \in \mathbb{R} \]

4. **Analyzing the Polynomial**:
The expression can be rearranged as:
\[ x^4 + (1-b)x^2 + b - bx^3 - bx > 0 \]
We can factor out the common terms:
\[ = x^2(x^2 + 1) - bx(x^2 + 1) + b \]
This simplifies to:
\[ (x^2 - bx + b)(x^2 + 1) > 0 \]

5. **Finding Conditions for Positivity**:
Since \( x^2 + 1 > 0 \) for all \( x \), we only need to ensure:
\[ x^2 - bx + b > 0 \quad \forall x \in \mathbb{R} \]
This is a quadratic in \( x \), and for it to be positive for all \( x \), its discriminant must be less than zero:
\[ D = b^2 - 4b < 0 \]
Simplifying gives:
\[ b(b - 4) < 0 \]

6. **Finding the Range of \( b \)**:
The inequality \( b(b - 4) < 0 \) implies:
- \( b < 0 \) or \( b > 4 \) gives positive values.
- \( 0 < b < 4 \) gives negative values.

Thus, \( b \) must lie in the interval:
\[ 0 < b < 4 \]

7. **Counting the Integer Values**:
The integers in the range \( (0, 4) \) are \( 1, 2, 3 \). Therefore, there are **3 integers** in this range.

### Final Answer:
The number of integers in the range of \( b \) is **3**.