If `I=int(x^(3)-1)/(x^(5)+x^(4)+x+1)dx=(1)/(4)ln(f(x))-ln(g(x))+c` (where, c is the constant of integration) and `f(0)=g(0)=1`,then the value of `f(1).g(1)` is equal to

Answer

To solve the problem, we need to find the functions \( f(x) \) and \( g(x) \) from the given integral expression and then evaluate \( f(1) \cdot g(1) \).

### Step-by-Step Solution:

1. **Identify the Integral**:
We start with the integral given:
\[ I = \int \frac{x^3 - 1}{x^5 + x^4 + x + 1} \, dx \]
This is expressed as:
\[ I = \frac{1}{4} \ln(f(x)) - \ln(g(x)) + C \]

2. **Simplify the Integral**:
We can rewrite the integrand:
\[ \frac{x^3 - 1}{x^5 + x^4 + x + 1} = \frac{x^3 - 1}{(x^4 + 1)(x + 1)} \]
This can be separated into two parts:
\[ I = \int \left( \frac{x^3}{x^5 + x^4 + x + 1} - \frac{1}{x + 1} \right) \, dx \]

3. **Substitution**:
Let \( t = x^4 + 1 \). Then, \( dt = 4x^3 \, dx \) or \( dx = \frac{dt}{4x^3} \).
The integral becomes:
\[ I_1 = \int \frac{x^3}{t} \cdot \frac{dt}{4x^3} = \frac{1}{4} \int \frac{1}{t} \, dt = \frac{1}{4} \ln |t| + C = \frac{1}{4} \ln |x^4 + 1| + C \]

4. **Integrate the Second Part**:
The second integral:
\[ I_2 = \int \frac{1}{x + 1} \, dx = \ln |x + 1| + C \]

5. **Combine Results**:
Combining both parts, we have:
\[ I = \frac{1}{4} \ln |x^4 + 1| - \ln |x + 1| + C \]

6. **Identify \( f(x) \) and \( g(x) \)**:
From the expression:
\[ \frac{1}{4} \ln(f(x)) - \ln(g(x)) = \frac{1}{4} \ln |x^4 + 1| - \ln |x + 1| \]
We can equate:
\[ f(x) = x^4 + 1 \quad \text{and} \quad g(x) = x + 1 \]

7. **Evaluate \( f(0) \) and \( g(0) \)**:
Check the initial conditions:
\[ f(0) = 0^4 + 1 = 1 \quad \text{and} \quad g(0) = 0 + 1 = 1 \]
Both conditions are satisfied.

8. **Calculate \( f(1) \) and \( g(1) \)**:
Now we compute:
\[ f(1) = 1^4 + 1 = 2 \quad \text{and} \quad g(1) = 1 + 1 = 2 \]

9. **Find \( f(1) \cdot g(1) \)**:
Finally, we find:
\[ f(1) \cdot g(1) = 2 \cdot 2 = 4 \]

### Final Answer:
The value of \( f(1) \cdot g(1) \) is \( 4 \).