Find the partial fraction
`(2x+1)/((3x+2)(4x^(2)+5x+6))`.

Answer

To find the partial fraction of the expression \(\frac{2x + 1}{(3x + 2)(4x^2 + 5x + 6)}\), we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We will express the given fraction as a sum of simpler fractions. Since the denominator consists of a linear factor \((3x + 2)\) and a quadratic factor \((4x^2 + 5x + 6)\), we can write:

\[ \frac{2x + 1}{(3x + 2)(4x^2 + 5x + 6)} = \frac{A}{3x + 2} + \frac{Bx + C}{4x^2 + 5x + 6} \]

where \(A\), \(B\), and \(C\) are constants to be determined.

### Step 2: Clear the denominators
Multiply both sides by the denominator \((3x + 2)(4x^2 + 5x + 6)\):

\[ 2x + 1 = A(4x^2 + 5x + 6) + (Bx + C)(3x + 2) \]

### Step 3: Expand the right-hand side
Now we will expand the right-hand side:

\[ A(4x^2 + 5x + 6) = 4Ax^2 + 5Ax + 6A \]
\[ (Bx + C)(3x + 2) = 3Bx^2 + 2Bx + 3Cx + 2C = 3Bx^2 + (2B + 3C)x + 2C \]

Combining these gives:

\[ 2x + 1 = (4A + 3B)x^2 + (5A + 2B + 3C)x + (6A + 2C) \]

### Step 4: Compare coefficients
Now we compare coefficients from both sides of the equation:

1. Coefficient of \(x^2\): \(4A + 3B = 0\) (Equation 1)
2. Coefficient of \(x\): \(5A + 2B + 3C = 2\) (Equation 2)
3. Constant term: \(6A + 2C = 1\) (Equation 3)

### Step 5: Solve the equations
From Equation 1, we can express \(B\) in terms of \(A\):

\[ B = -\frac{4A}{3} \]

Substituting \(B\) into Equation 2:

\[ 5A + 2\left(-\frac{4A}{3}\right) + 3C = 2 \]
\[ 5A - \frac{8A}{3} + 3C = 2 \]
Multiplying through by 3 to eliminate the fraction:

\[ 15A - 8A + 9C = 6 \]
\[ 7A + 9C = 6 \quad \text{(Equation 4)} \]

Now substituting \(B\) into Equation 3:

\[ 6A + 2C = 1 \]

### Step 6: Solve Equations 4 and 5
From Equation 3, we can express \(C\) in terms of \(A\):

\[ 2C = 1 - 6A \implies C = \frac{1 - 6A}{2} \]

Substituting \(C\) into Equation 4:

\[ 7A + 9\left(\frac{1 - 6A}{2}\right) = 6 \]
Multiplying through by 2:

\[ 14A + 9 - 54A = 12 \]
\[ -40A = 3 \implies A = -\frac{3}{40} \]

Now substituting \(A\) back to find \(B\) and \(C\):

\[ B = -\frac{4(-\frac{3}{40})}{3} = \frac{12}{120} = \frac{1}{10} \]
\[ C = \frac{1 - 6(-\frac{3}{40})}{2} = \frac{1 + \frac{18}{40}}{2} = \frac{\frac{40 + 18}{40}}{2} = \frac{58/40}{2} = \frac{29}{40} \]

### Step 7: Write the final partial fraction decomposition
Now we can write the partial fraction decomposition:

\[ \frac{2x + 1}{(3x + 2)(4x^2 + 5x + 6)} = \frac{-\frac{3}{40}}{3x + 2} + \frac{\frac{1}{10}x + \frac{29}{40}}{4x^2 + 5x + 6} \]

### Final Answer:
\[ \frac{2x + 1}{(3x + 2)(4x^2 + 5x + 6)} = \frac{-\frac{3}{40}}{3x + 2} + \frac{\frac{1}{10}x + \frac{29}{40}}{4x^2 + 5x + 6} \]