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 can follow these steps:

### Step 1: Set up the partial fraction decomposition
We can express the given fraction as:
\[ \frac{2x+1}{(3x+2)(4x^2+5x+6)} = \frac{A}{3x+2} + \frac{Bx+C}{4x^2+5x+6} \]
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
Expanding the right-hand side gives:
\[ 2x + 1 = A(4x^2 + 5x + 6) + (3Bx^2 + 2Bx + 3Cx + 2C) \]
Combining like terms:
\[ 2x + 1 = (4A + 3B)x^2 + (5A + 2B + 3C)x + (6A + 2C) \]

### Step 4: Set up a system of equations
Now, we can equate the coefficients from both sides:
1. For \(x^2\): \(4A + 3B = 0\) (1)
2. For \(x\): \(5A + 2B + 3C = 2\) (2)
3. For the constant term: \(6A + 2C = 1\) (3)

### Step 5: Solve the system of equations
From equation (1):
\[ B = -\frac{4A}{3} \]

Substituting \(B\) into equations (2) and (3):
1. Substitute into (2):
\[ 5A + 2\left(-\frac{4A}{3}\right) + 3C = 2 \]
This simplifies to:
\[ 5A - \frac{8A}{3} + 3C = 2 \]
Multiplying through by 3 to eliminate the fraction:
\[ 15A - 8A + 9C = 6 \implies 7A + 9C = 6 \quad (4) \]

2. Substitute into (3):
\[ 6A + 2C = 1 \quad (5) \]

Now we have a new system of equations (4) and (5):
1. \(7A + 9C = 6\) (4)
2. \(6A + 2C = 1\) (5)

### Step 6: Solve for \(C\) in terms of \(A\)
From equation (5):
\[ 2C = 1 - 6A \implies C = \frac{1 - 6A}{2} \]

### Step 7: Substitute \(C\) into equation (4)
Substituting \(C\) into (4):
\[ 7A + 9\left(\frac{1 - 6A}{2}\right) = 6 \]
Multiplying through by 2:
\[ 14A + 9 - 54A = 12 \implies -40A = 3 \implies A = -\frac{3}{40} \]

### Step 8: Find \(B\) and \(C\)
Substituting \(A\) back to find \(B\):
\[ B = -\frac{4(-\frac{3}{40})}{3} = \frac{12}{120} = \frac{1}{10} \]
Now substituting \(A\) into the equation for \(C\):
\[ C = \frac{1 - 6(-\frac{3}{40})}{2} = \frac{1 + \frac{18}{40}}{2} = \frac{\frac{40 + 18}{40}}{2} = \frac{58}{80} = \frac{29}{40} \]

### Final Result
Thus, the partial fraction decomposition is:
\[ \frac{2x+1}{(3x+2)(4x^2+5x+6)} = \frac{-\frac{3}{40}}{3x+2} + \frac{\frac{1}{10}x + \frac{29}{40}}{4x^2 + 5x + 6} \]