If `int (3sin x+2 cosx)/(3cosx+2sinx)dx=ax+blog_(e)|2 sinx+3 cosx|+c` then

Options

A. `a= -(12)/(13)`
B. `b=(6)/(13)`
C. `a=(12)/(13)`
D. `b= -(15)/(39)`

Explanation

To solve the integral equation given in the question, we will differentiate both sides with respect to \( x \) and then equate the coefficients of the resulting expressions.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \, dx = ax + b \log_e |2 \sin x + 3 \cos x| + c \]

2. **Differentiate both sides with respect to \( x \):**
\[ \frac{d}{dx} \left( \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \, dx \right) = \frac{d}{dx} \left( ax + b \log_e |2 \sin x + 3 \cos x| + c \right) \]

3. **Apply the Fundamental Theorem of Calculus on the left side:**
\[ \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \]

4. **Differentiate the right side:**
- The derivative of \( ax \) is \( a \).
- The derivative of \( b \log_e |2 \sin x + 3 \cos x| \) using the chain rule is:
\[ b \cdot \frac{1}{2 \sin x + 3 \cos x} \cdot (2 \cos x - 3 \sin x) \]

Therefore, the right side becomes:
\[ a + b \cdot \frac{2 \cos x - 3 \sin x}{2 \sin x + 3 \cos x} \]

5. **Set both sides equal:**
\[ \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} = a + b \cdot \frac{2 \cos x - 3 \sin x}{2 \sin x + 3 \cos x} \]

6. **Multiply both sides by \( (3 \cos x + 2 \sin x)(2 \sin x + 3 \cos x) \) to eliminate the denominator:**
\[ (3 \sin x + 2 \cos x)(2 \sin x + 3 \cos x) = (a(3 \cos x + 2 \sin x)(2 \sin x + 3 \cos x) + b(2 \cos x - 3 \sin x)(3 \cos x + 2 \sin x)) \]

7. **Expand both sides and collect like terms:**
- Left side:
\[ 6 \sin^2 x + 9 \sin x \cos x + 4 \cos^2 x \]
- Right side:
\[ 2a \sin x + 3a \cos x + 2b \cos x - 3b \sin x \]

8. **Equate coefficients of \( \sin x \) and \( \cos x \):**
- For \( \sin x \):
\[ 2a - 3b = 3 \]
- For \( \cos x \):
\[ 3a + 2b = 2 \]

9. **Solve the system of equations:**
- From \( 2a - 3b = 3 \) (1)
- From \( 3a + 2b = 2 \) (2)

Multiply (1) by 3:
\[ 6a - 9b = 9 \]

Multiply (2) by 2:
\[ 6a + 4b = 4 \]

Subtract the second from the first:
\[ -13b = 5 \implies b = -\frac{5}{13} \]

Substitute \( b \) back into (1):
\[ 2a - 3(-\frac{5}{13}) = 3 \implies 2a + \frac{15}{13} = 3 \]
\[ 2a = 3 - \frac{15}{13} = \frac{39 - 15}{13} = \frac{24}{13} \implies a = \frac{12}{13} \]

10. **Final values:**
\[ a = \frac{12}{13}, \quad b = -\frac{5}{13} \]