The value of `int(3x+2)/((x-2)^(2)(x-3))dx` is equal to

Options

A. `11log (x-3)/(x-2)-(8)/(x-2)+C`
B. `11log (x+3)/(x+2)-(8)/(x-2)+C`
C. `11log(x-3)/(x-2)+(8)/(x-2)+C`
D. `11log (x+3)/(x+2)+(8)/(x-2)+C`

Explanation

To solve the integral \( I = \int \frac{3x + 2}{(x - 2)^2 (x - 3)} \, dx \), we will use the method of partial fractions.

### Step-by-step Solution:

1. **Set up the partial fraction decomposition**:
We want to express the integrand as:
\[ \frac{3x + 2}{(x - 2)^2 (x - 3)} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} + \frac{C}{x - 3} \]
where \( A \), \( B \), and \( C \) are constants to be determined.

2. **Multiply through by the denominator**:
Multiply both sides by \( (x - 2)^2 (x - 3) \):
\[ 3x + 2 = A(x - 2)(x - 3) + B(x - 3) + C(x - 2)^2 \]

3. **Expand the right-hand side**:
Expanding gives:
\[ A(x^2 - 5x + 6) + B(x - 3) + C(x^2 - 4x + 4) \]
Combining like terms, we get:
\[ (A + C)x^2 + (-5A - 4C + B)x + (6A - 3B + 4C) \]

4. **Set up equations for coefficients**:
Now, we equate coefficients from both sides:
- For \( x^2 \): \( A + C = 0 \) (1)
- For \( x \): \( -5A - 4C + B = 3 \) (2)
- For the constant term: \( 6A - 3B + 4C = 2 \) (3)

5. **Solve the system of equations**:
From equation (1): \( C = -A \).
Substitute \( C \) into equations (2) and (3):
- Equation (2):
\[ -5A - 4(-A) + B = 3 \implies -5A + 4A + B = 3 \implies -A + B = 3 \implies B = A + 3 \quad (4) \]
- Equation (3):
\[ 6A - 3(A + 3) + 4(-A) = 2 \implies 6A - 3A - 9 - 4A = 2 \implies -A - 9 = 2 \implies -A = 11 \implies A = -11 \]
Now substituting \( A \) back into equations (1) and (4):
- From (1): \( C = -(-11) = 11 \)
- From (4): \( B = -11 + 3 = -8 \)

6. **Substituting back into the partial fractions**:
We have:
\[ \frac{3x + 2}{(x - 2)^2 (x - 3)} = \frac{-11}{x - 2} + \frac{-8}{(x - 2)^2} + \frac{11}{x - 3} \]

7. **Integrate each term**:
Now we can integrate term by term:
\[ I = \int \left( \frac{-11}{x - 2} + \frac{-8}{(x - 2)^2} + \frac{11}{x - 3} \right) dx \]
- The first integral:
\[ \int \frac{-11}{x - 2} \, dx = -11 \ln |x - 2| \]
- The second integral:
\[ \int \frac{-8}{(x - 2)^2} \, dx = 8 \cdot \frac{1}{x - 2} \quad \text{(since } \int x^{-2} \, dx = -x^{-1} \text{)} \]
- The third integral:
\[ \int \frac{11}{x - 3} \, dx = 11 \ln |x - 3| \]

8. **Combine the results**:
Putting it all together, we get:
\[ I = -11 \ln |x - 2| + \frac{8}{x - 2} + 11 \ln |x - 3| + C \]
where \( C \) is the constant of integration.

### Final Answer:
\[ I = -11 \ln |x - 2| + \frac{8}{x - 2} + 11 \ln |x - 3| + C \]