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

Answer

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

### Step 1: Set up the partial fraction decomposition
We start by expressing the given fraction as a sum of simpler fractions. Since we have a repeated linear factor \((1-2x)^2\) and a simple linear factor \((1-3x)\), we can write:

\[ \frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-2x} + \frac{B}{(1-2x)^2} + \frac{C}{1-3x} \]

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

### Step 2: Clear the denominators
Multiply both sides by the common denominator \((1-2x)^2(1-3x)\) to eliminate the fractions:

\[ 1 = A(1-2x)(1-3x) + B(1-3x) + C(1-2x)^2 \]

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

1. Expand \(A(1-2x)(1-3x)\):
\[ A(1 - 2x - 3x + 6x^2) = A(1 - 5x + 6x^2) \]

2. Expand \(B(1-3x)\):
\[ B(1 - 3x) \]

3. Expand \(C(1-2x)^2\):
\[ C(1 - 4x + 4x^2) \]

Combining these gives:
\[ 1 = A(1 - 5x + 6x^2) + B(1 - 3x) + C(1 - 4x + 4x^2) \]

### Step 4: Combine like terms
Now, we combine all the terms on the right-hand side:

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

### Step 5: Set up equations
Now we can equate the coefficients from both sides of the equation. The left side has coefficients:
- Constant term: \(1\)
- Coefficient of \(x\): \(0\)
- Coefficient of \(x^2\): \(0\)

From this, we get the following system of equations:
1. \(A + B + C = 1\)
2. \(-5A - 3B - 4C = 0\)
3. \(6A + 4C = 0\)

### Step 6: Solve the system of equations
From the third equation \(6A + 4C = 0\), we can express \(C\) in terms of \(A\):
\[ C = -\frac{3}{2}A \]

Substituting \(C\) into the first equation:
\[ A + B - \frac{3}{2}A = 1 \implies -\frac{1}{2}A + B = 1 \implies B = 1 + \frac{1}{2}A \]

Now substitute \(B\) and \(C\) into the second equation:
\[ -5A - 3(1 + \frac{1}{2}A) - 4(-\frac{3}{2}A) = 0 \]
\[ -5A - 3 - \frac{3}{2}A + 6A = 0 \]
\[ \frac{1}{2}A - 3 = 0 \implies A = 6 \]

Now substitute \(A\) back to find \(B\) and \(C\):
\[ C = -\frac{3}{2}(6) = -9 \]
\[ B = 1 + \frac{1}{2}(6) = 4 \]

### Step 7: Write the partial fraction decomposition
Now we can write the partial fraction decomposition:
\[ \frac{1}{(1-2x)^2(1-3x)} = \frac{6}{1-2x} + \frac{4}{(1-2x)^2} - \frac{9}{1-3x} \]

### Final Answer:
Thus, the partial fraction decomposition of \(\frac{1}{(1-2x)^2(1-3x)}\) is:
\[ \frac{6}{1-2x} + \frac{4}{(1-2x)^2} - \frac{9}{1-3x} \]

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