`int(x^2)/((x-1)(3x-1)(3x-2))dx=`

Options

A. `(1)/(2)log|x-1|+(1)/(18)log|3x-1|+(4)/(9)log|3x-2|+c`
B. `(1)/(2)log|x-1|-(1)/(18)log|3x-1|+(4)/(9)log|3x-2|+c`
C. `(1)/(2)log|x-1|+(1)/(18)log|3x-1|-(4)/(9)log|3x-2|+c`
D. `(1)/(2)log|x-1|-(1)/(18)log|3x-1|-(4)/(9)log|3x-2|+c`

Answer

To solve the integral

\[ \int \frac{x^2}{(x-1)(3x-1)(3x-2)} \, dx, \]

we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Set up the partial fraction decomposition

We can express the integrand as:

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

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

### Step 2: Multiply through by the denominator

Multiplying both sides by the denominator \((x-1)(3x-1)(3x-2)\) gives:

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

### Step 3: Expand the right-hand side

Now we expand the right-hand side:

1. For \(A(3x-1)(3x-2)\):
\[ A(9x^2 - 15x + 2) \]

2. For \(B(x-1)(3x-2)\):
\[ B(3x^2 - 5x + 2) \]

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

Combining these, we get:

\[ x^2 = (9A + 3B + 3C)x^2 + (-15A - 5B - 4C)x + (2A + 2B + C) \]

### Step 4: Set up the system of equations

Now, we equate coefficients from both sides:

1. For \(x^2\): \(9A + 3B + 3C = 1\)
2. For \(x\): \(-15A - 5B - 4C = 0\)
3. For the constant term: \(2A + 2B + C = 0\)

### Step 5: Solve the system of equations

From the third equation, we can express \(C\) in terms of \(A\) and \(B\):

\[ C = -2A - 2B \]

Substituting \(C\) into the first two equations:

1. \(9A + 3B + 3(-2A - 2B) = 1\)
\[ 9A + 3B - 6A - 6B = 1 \implies 3A - 3B = 1 \implies A - B = \frac{1}{3} \quad (i) \]

2. \(-15A - 5B - 4(-2A - 2B) = 0\)
\[ -15A - 5B + 8A + 8B = 0 \implies -7A + 3B = 0 \implies 3B = 7A \implies B = \frac{7}{3}A \quad (ii) \]

Substituting \(B\) from equation (ii) into (i):

\[ A - \frac{7}{3}A = \frac{1}{3} \implies -\frac{4}{3}A = \frac{1}{3} \implies A = -\frac{1}{4} \]

Now substituting \(A\) back into (ii):

\[ B = \frac{7}{3} \left(-\frac{1}{4}\right) = -\frac{7}{12} \]

And substituting \(A\) and \(B\) into the expression for \(C\):

\[ C = -2\left(-\frac{1}{4}\right) - 2\left(-\frac{7}{12}\right) = \frac{1}{2} + \frac{7}{6} = \frac{3}{6} + \frac{7}{6} = \frac{10}{6} = \frac{5}{3} \]

### Step 6: Write the partial fraction decomposition

Thus, we have:

\[ \frac{x^2}{(x-1)(3x-1)(3x-2)} = \frac{-\frac{1}{4}}{x-1} + \frac{-\frac{7}{12}}{3x-1} + \frac{\frac{5}{3}}{3x-2} \]

### Step 7: Integrate each term

Now we can integrate each term separately:

\[ \int \left( \frac{-\frac{1}{4}}{x-1} + \frac{-\frac{7}{12}}{3x-1} + \frac{\frac{5}{3}}{3x-2} \right) dx \]

This gives:

\[ -\frac{1}{4} \ln |x-1| - \frac{7}{36} \ln |3x-1| + \frac{5}{9} \ln |3x-2| + C \]

### Final Answer

Thus, the final answer is:

\[ -\frac{1}{4} \ln |x-1| - \frac{7}{36} \ln |3x-1| + \frac{5}{9} \ln |3x-2| + C \]