Let `f (x) = int _(0)^(x) ((a -1) (t ^(2)+t+1)^(2) -(a+1)(t^(4)+t ^(2) +1))` dt. Then the total number of integral values of 'a' for which `f'(x)=0` has no real roots is

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined by the integral and find the conditions under which its derivative \( f'(x) \) has no real roots.

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \int_{0}^{x} \left((a - 1)(t^2 + t + 1)^2 - (a + 1)(t^4 + t^2 + 1)\right) dt \]

2. **Find the derivative \( f'(x) \)** using the Fundamental Theorem of Calculus:
\[ f'(x) = (a - 1)(x^2 + x + 1)^2 - (a + 1)(x^4 + x^2 + 1) \]

3. **Set \( f'(x) = 0 \)** to find the roots:
\[ (a - 1)(x^2 + x + 1)^2 = (a + 1)(x^4 + x^2 + 1) \]

4. **Rearranging the equation** gives:
\[ \frac{a - 1}{a + 1} = \frac{x^4 + x^2 + 1}{(x^2 + x + 1)^2} \]

5. **Let \( k = \frac{a - 1}{a + 1} \)**, then we can express \( a \) in terms of \( k \):
\[ a = \frac{2k + 1}{1 - k} \]

6. **Analyze the right-hand side**:
\[ \frac{x^4 + x^2 + 1}{(x^2 + x + 1)^2} \]
This is a rational function. We need to analyze its behavior to find conditions for \( k \).

7. **Find the discriminant of the quadratic equation** derived from setting \( f'(x) = 0 \):
The equation can be expressed as:
\[ x^2 - ax + 1 = 0 \]
The discriminant \( D \) must be less than 0 for the equation to have no real roots:
\[ D = a^2 - 4 < 0 \implies a^2 < 4 \]

8. **Solve the inequality**:
\[ -2 < a < 2 \]

9. **Identify integral values of \( a \)** within the interval:
The integral values of \( a \) in the interval \( (-2, 2) \) are:
\[ -1, 0, 1 \]

10. **Count the integral values**:
There are **3 integral values** of \( a \) that satisfy the condition.

### Final Answer:
The total number of integral values of \( a \) for which \( f'(x) = 0 \) has no real roots is **3**.