Evaluate the following Integrals :
`int (x^(2))/((x-1)(x-2)(x-3))dx`

Answer

To evaluate the integral

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

we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition
We can express the integrand as:

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

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

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

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

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

Combining these gives:

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

### Step 4: Set up the system of equations
Now, we can equate coefficients from both sides:

1. \(A + B + C = 1\) (coefficient of \(x^2\))
2. \(-5A - 4B - 3C = 0\) (coefficient of \(x\))
3. \(6A + 3B + 2C = 0\) (constant term)

### Step 5: Solve the system of equations
From equation (1), we have:

\[ C = 1 - A - B. \]

Substituting \(C\) into equations (2) and (3):

\[ -5A - 4B - 3(1 - A - B) = 0 \implies -5A - 4B - 3 + 3A + 3B = 0 \implies -2A - B = 3 \quad (4) \]

\[ 6A + 3B + 2(1 - A - B) = 0 \implies 6A + 3B + 2 - 2A - 2B = 0 \implies 4A + B + 2 = 0 \quad (5) \]

Now we can solve equations (4) and (5):

From (4):

\[ B = -2A - 3. \]

Substituting \(B\) into (5):

\[ 4A + (-2A - 3) + 2 = 0 \implies 2A - 1 = 0 \implies A = \frac{1}{2}. \]

Substituting \(A\) back into (4):

\[ B = -2\left(\frac{1}{2}\right) - 3 = -1 - 3 = -4. \]

Now substituting \(A\) and \(B\) into (1):

\[ C = 1 - \frac{1}{2} + 4 = \frac{1}{2} + 4 = \frac{9}{2}. \]

### Step 6: Write the partial fraction decomposition
Thus, we have:

\[ \frac{x^2}{(x-1)(x-2)(x-3)} = \frac{1/2}{x-1} - \frac{4}{x-2} + \frac{9/2}{x-3}. \]

### Step 7: Integrate each term
Now we can integrate:

\[ \int \left( \frac{1/2}{x-1} - \frac{4}{x-2} + \frac{9/2}{x-3} \right) dx = \frac{1}{2} \ln |x-1| - 4 \ln |x-2| + \frac{9}{2} \ln |x-3| + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ \int \frac{x^2}{(x-1)(x-2)(x-3)} \, dx = \frac{1}{2} \ln |x-1| - 4 \ln |x-2| + \frac{9}{2} \ln |x-3| + C. \]