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 \( ax^2 + bx + c \). Let's break down the solution step by step.

### Step 1: Set up the integral
We start with the integral:
\[ \int \frac{(2x + 3) \, dx}{x(x + 1)(x + 2)(x + 3) + 1} \]

### Step 2: Simplify the denominator
First, we need to simplify the denominator:
\[ x(x + 1)(x + 2)(x + 3) + 1 \]
Expanding \( x(x + 1)(x + 2)(x + 3) \):
\[ = x[(x^2 + 3x + 2)(x + 3)] = x[x^3 + 6x^2 + 11x + 6] \]
Thus,
\[ = x^4 + 6x^3 + 11x^2 + 6x + 1 \]

### Step 3: Change of variables
We notice that the numerator \( 2x + 3 \) can be related to the derivative of \( x^2 + 3x \). Let's set:
\[ t = x^2 + 3x \]
Then, the derivative \( dt = (2x + 3) \, dx \).

### Step 4: Rewrite the integral
Now, we can rewrite the integral in terms of \( t \):
\[ \int \frac{dt}{t(t + 2) + 1} \]
This simplifies to:
\[ \int \frac{dt}{t^2 + 2t + 1} = \int \frac{dt}{(t + 1)^2} \]

### Step 5: Integrate
The integral of \( \frac{1}{(t + 1)^2} \) is:
\[ -\frac{1}{t + 1} + C \]

### Step 6: Substitute back for \( t \)
Substituting back for \( t \):
\[ -\frac{1}{x^2 + 3x + 1} + C \]

### Step 7: Identify \( f(x) \)
From the problem statement, we have:
\[ C - \frac{1}{f(x)} = -\frac{1}{x^2 + 3x + 1} + C \]
This implies:
\[ f(x) = x^2 + 3x + 1 \]

### Step 8: Calculate \( f(1) \)
Now, we need to find \( f(1) \):
\[ f(1) = 1^2 + 3(1) + 1 = 1 + 3 + 1 = 5 \]

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