Resolve `(1)/((x-1)(x+2)(2x+3))` into partial fractions.

Answer

To resolve the expression \( \frac{1}{(x-1)(x+2)(2x+3)} \) into partial fractions, we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We express the given fraction as a sum of simpler fractions:
\[ \frac{1}{(x-1)(x+2)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{2x+3} \]
where \( A \), \( B \), and \( C \) are constants that we need to determine.

### Step 2: Clear the denominators
Multiply both sides by the denominator \( (x-1)(x+2)(2x+3) \):
\[ 1 = A(x+2)(2x+3) + B(x-1)(2x+3) + C(x-1)(x+2) \]

### Step 3: Expand the right-hand side
Now we will expand the right-hand side:
1. For \( A(x+2)(2x+3) \):
\[ A(2x^2 + 3x + 4x + 6) = A(2x^2 + 7x + 6) \]
2. For \( B(x-1)(2x+3) \):
\[ B(2x^2 + 3x - 2x - 3) = B(2x^2 + x - 3) \]
3. For \( C(x-1)(x+2) \):
\[ C(x^2 + 2x - x - 2) = C(x^2 + x - 2) \]

Combining these, we have:
\[ 1 = (2A + 2B + C)x^2 + (7A + B + C)x + (6A - 3B - 2C) \]

### Step 4: Set up equations
Now, we equate the coefficients from both sides:
1. Coefficient of \( x^2 \): \( 2A + 2B + C = 0 \)
2. Coefficient of \( x \): \( 7A + B + C = 0 \)
3. Constant term: \( 6A - 3B - 2C = 1 \)

### Step 5: Solve the system of equations
We will solve these equations step by step.

1. From the first equation, express \( C \):
\[ C = -2A - 2B \]

2. Substitute \( C \) into the second equation:
\[ 7A + B - 2A - 2B = 0 \implies 5A - B = 0 \implies B = 5A \]

3. Substitute \( B \) and \( C \) into the third equation:
\[ 6A - 3(5A) - 2(-2A - 2(5A)) = 1 \]
Simplifying gives:
\[ 6A - 15A + 4A + 20A = 1 \implies 15A = 1 \implies A = \frac{1}{15} \]

4. Now substitute \( A \) back to find \( B \) and \( C \):
\[ B = 5A = 5 \cdot \frac{1}{15} = \frac{1}{3} \]
\[ C = -2A - 2B = -2 \cdot \frac{1}{15} - 2 \cdot \frac{1}{3} = -\frac{2}{15} - \frac{10}{15} = -\frac{12}{15} = -\frac{4}{5} \]

### Step 6: Write the final result
Now substituting \( A \), \( B \), and \( C \) back into the partial fractions:
\[ \frac{1}{(x-1)(x+2)(2x+3)} = \frac{1/15}{x-1} + \frac{1/3}{x+2} - \frac{4/5}{2x+3} \]

### Final Answer
Thus, the resolved partial fraction is:
\[ \frac{1}{15(x-1)} + \frac{1}{3(x+2)} - \frac{4/5}{2x+3} \]