If `int((3x+4))/((x^(3)-2x-4))dx=Alog|x-2|+Blog(f(x))+c`, then

Options

A. `B=-(1)/(2)`
B. A=1
C. `f(x)=|x^(2)+2x+2|`
D. `B=-(1)/(4)`

Answer

To solve the integral \( \int \frac{3x + 4}{x^3 - 2x - 4} \, dx \) and express it in the form \( A \log |x - 2| + B \log(f(x)) + C \), we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \( x^3 - 2x - 4 \). We will check for possible rational roots using the Rational Root Theorem.

1. **Check for roots**: Let's test \( x = 2 \):
\[ 2^3 - 2(2) - 4 = 8 - 4 - 4 = 0 \]
Thus, \( x - 2 \) is a factor.

2. **Perform polynomial long division** to divide \( x^3 - 2x - 4 \) by \( x - 2 \):
\[ x^3 - 2x - 4 = (x - 2)(x^2 + 2x + 2) \]

### Step 2: Set Up Partial Fraction Decomposition
Now we can express the integrand using partial fractions:
\[ \frac{3x + 4}{(x - 2)(x^2 + 2x + 2)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + 2x + 2} \]

### Step 3: Solve for Coefficients A, B, and C
Multiply through by the denominator to eliminate the fractions:
\[ 3x + 4 = A(x^2 + 2x + 2) + (Bx + C)(x - 2) \]

Expanding the right side:
\[ 3x + 4 = Ax^2 + 2Ax + 2A + Bx^2 - 2Bx + Cx - 2C \]
Combine like terms:
\[ 3x + 4 = (A + B)x^2 + (2A - 2B + C)x + (2A - 2C) \]

Now, equate coefficients:
1. \( A + B = 0 \) (coefficient of \( x^2 \))
2. \( 2A - 2B + C = 3 \) (coefficient of \( x \))
3. \( 2A - 2C = 4 \) (constant term)

### Step 4: Solve the System of Equations
From \( A + B = 0 \), we have \( B = -A \).

Substituting \( B = -A \) into the other equations:
1. \( 2A - 2(-A) + C = 3 \) simplifies to \( 4A + C = 3 \)
2. \( 2A - 2C = 4 \)

Now we have:
1. \( C = 3 - 4A \)
2. Substitute \( C \) into \( 2A - 2(3 - 4A) = 4 \):
\[ 2A - 6 + 8A = 4 \implies 10A = 10 \implies A = 1 \]
Thus, \( B = -1 \) and \( C = -1 \).

### Step 5: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left( \frac{1}{x - 2} - \frac{x + 1}{x^2 + 2x + 2} \right) dx \]

### Step 6: Integrate Each Term
1. For \( \frac{1}{x - 2} \):
\[ \int \frac{1}{x - 2} \, dx = \log |x - 2| \]

2. For \( \frac{x + 1}{x^2 + 2x + 2} \), use substitution:
Let \( u = x^2 + 2x + 2 \), then \( du = (2x + 2)dx \) or \( dx = \frac{du}{2(x + 1)} \):
\[ \int \frac{x + 1}{x^2 + 2x + 2} \, dx = \frac{1}{2} \log |x^2 + 2x + 2| \]

### Step 7: Combine Results
Thus, the integral becomes:
\[ \log |x - 2| - \frac{1}{2} \log |x^2 + 2x + 2| + C \]

### Final Form
This can be rewritten as:
\[ \log |x - 2| + \log |(x^2 + 2x + 2)^{-1/2}| + C \]

### Conclusion
Comparing with the given form \( A \log |x - 2| + B \log(f(x)) + C \):
- \( A = 1 \)
- \( B = -\frac{1}{2} \)
- \( f(x) = |x^2 + 2x + 2| \)