If `int((2x+3)dx)/(x(x+1)(x+2)(x+3)+1)=C-(1)/(f(x))` where f(x) is of the form of `ax^(2)+bx+c`, then the value of f(1) is

Options

A. 4
B. 5
C. 6
D. none

Explanation

To solve the given problem, we need to evaluate the integral and find the function \( f(x) \) in the form \( ax^2 + bx + c \) such that \( f(1) \) can be computed.

### Step-by-Step Solution:

1. **Set Up the Integral**:
We start with the integral:
\[ I = \int \frac{2x + 3}{x(x + 1)(x + 2)(x + 3) + 1} \, dx \]
We know that this integral is equal to:
\[ C - \frac{1}{f(x)} \]
where \( f(x) \) is of the form \( ax^2 + bx + c \).

2. **Substitution**:
Let’s define \( t = x^2 + 3x \). Then, the derivative \( dt \) is:
\[ dt = (2x + 3) \, dx \]
This allows us to rewrite the integral in terms of \( t \):
\[ I = \int \frac{dt}{t(t + 2) + 1} \]

3. **Simplifying the Denominator**:
We simplify the denominator:
\[ t(t + 2) + 1 = t^2 + 2t + 1 = (t + 1)^2 \]
Thus, the integral becomes:
\[ I = \int \frac{dt}{(t + 1)^2} \]

4. **Integrating**:
The integral of \( \frac{1}{(t + 1)^2} \) is:
\[ -\frac{1}{t + 1} + C \]

5. **Substituting Back**:
Now we substitute back \( t = x^2 + 3x \):
\[ I = -\frac{1}{x^2 + 3x + 1} + C \]

6. **Comparing with Given Expression**:
We know from the problem statement that:
\[ I = C - \frac{1}{f(x)} \]
Therefore, we can compare:
\[ -\frac{1}{x^2 + 3x + 1} + C = C - \frac{1}{f(x)} \]
This implies:
\[ f(x) = x^2 + 3x + 1 \]

7. **Finding \( f(1) \)**:
Now we need to find \( f(1) \):
\[ f(1) = 1^2 + 3 \cdot 1 + 1 = 1 + 3 + 1 = 5 \]

### Final Answer:
Thus, the value of \( f(1) \) is:
\[ \boxed{5} \]