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 \( \int \frac{3x + 4}{x^3 - 2x - 4} \, dx \), we will follow these steps:

### Step 1: Factor the Denominator
We need to factor the polynomial in the denominator \( x^3 - 2x - 4 \). We can use the Rational Root Theorem or trial and error to find a root.

By substituting \( x = 2 \):
\[ 2^3 - 2(2) - 4 = 8 - 4 - 4 = 0 \]
Thus, \( x - 2 \) is a factor. We can perform polynomial long division or synthetic division to factor the cubic polynomial.

### Step 2: Polynomial Long Division
Dividing \( x^3 - 2x - 4 \) by \( x - 2 \):
1. Divide the leading term: \( x^3 / x = x^2 \)
2. Multiply \( x^2 \) by \( x - 2 \): \( x^3 - 2x^2 \)
3. Subtract: \( (x^3 - 2x - 4) - (x^3 - 2x^2) = 2x^2 - 2x - 4 \)
4. Repeat: Divide \( 2x^2 / x = 2x \)
5. Multiply: \( 2x(x - 2) = 2x^2 - 4x \)
6. Subtract: \( (2x^2 - 2x - 4) - (2x^2 - 4x) = 2x - 4 \)
7. Repeat: Divide \( 2x / x = 2 \)
8. Multiply: \( 2(x - 2) = 2x - 4 \)
9. Subtract: \( (2x - 4) - (2x - 4) = 0 \)

Thus, we have:
\[ x^3 - 2x - 4 = (x - 2)(x^2 + 2x + 2) \]

### Step 3: Partial Fraction Decomposition
We can express:
\[ \frac{3x + 4}{(x - 2)(x^2 + 2x + 2)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + 2x + 2} \]
Multiplying through by the denominator:
\[ 3x + 4 = A(x^2 + 2x + 2) + (Bx + C)(x - 2) \]

### Step 4: Expand and Collect Terms
Expanding:
\[ 3x + 4 = A(x^2 + 2x + 2) + Bx^2 - 2Bx + Cx - 2C \]
Combining like terms:
\[ 3x + 4 = (A + B)x^2 + (2A - 2B + C)x + (2A - 2C) \]

### Step 5: Set Up the System of Equations
Equating coefficients:
1. \( A + B = 0 \) (coefficient of \( x^2 \))
2. \( 2A - 2B + C = 3 \) (coefficient of \( x \))
3. \( 2A - 2C = 4 \) (constant term)

### Step 6: Solve the System
From \( A + B = 0 \), we have \( B = -A \). Substituting into the other equations:
1. \( 2A - 2(-A) + C = 3 \) becomes \( 4A + C = 3 \)
2. \( 2A - 2C = 4 \)

From the second equation, \( C = A - 2 \). Substituting this into \( 4A + (A - 2) = 3 \):
\[ 5A - 2 = 3 \implies 5A = 5 \implies A = 1 \]
Thus, \( B = -1 \) and \( C = -1 \).

### Step 7: 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 8: Integrate Each Term
1. \( \int \frac{1}{x - 2} \, dx = \ln |x - 2| \)
2. For \( \int \frac{x + 1}{x^2 + 2x + 2} \, dx \), we use substitution \( u = x^2 + 2x + 2 \), \( du = (2x + 2)dx \), leading to:
\[ \int \frac{1}{2} \frac{du}{u} = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x^2 + 2x + 2| \]

### Final Answer
Combining the results:
\[ \int \frac{3x + 4}{x^3 - 2x - 4} \, dx = \ln |x - 2| - \frac{1}{2} \ln |x^2 + 2x + 2| + C \]