`int (1)/((x-2)(x+2)(x-2))dx`

Answer

To solve the integral \( \int \frac{1}{(x-2)(x+2)(x-2)} \, dx \), we can follow these steps:

### Step 1: Simplify the Integral
We can rewrite the integral as:
\[ \int \frac{1}{(x-2)^2 (x+2)} \, dx \]

### Step 2: Set Up Partial Fractions
We will express the integrand using partial fractions:
\[ \frac{1}{(x-2)^2 (x+2)} = \frac{A}{x+2} + \frac{B}{x-2} + \frac{C}{(x-2)^2} \]
where \( A \), \( B \), and \( C \) are constants to be determined.

### Step 3: Combine the Right Side
Multiplying both sides by the denominator \((x-2)^2 (x+2)\) gives:
\[ 1 = A(x-2)^2 + B(x-2)(x+2) + C(x+2) \]

### Step 4: Expand the Right Side
Expanding the right-hand side:
\[ 1 = A(x^2 - 4x + 4) + B(x^2 - 4) + C(x + 2) \]
\[ = (A + B)x^2 + (-4A + C)x + (4A - 4B + 2C) \]

### Step 5: Set Up the System of Equations
Now, we can equate coefficients from both sides:
1. \( A + B = 0 \) (coefficient of \( x^2 \))
2. \( -4A + C = 0 \) (coefficient of \( x \))
3. \( 4A - 4B + 2C = 1 \) (constant term)

### Step 6: Solve the System of Equations
From equation (1), we can express \( B \) in terms of \( A \):
\[ B = -A \]

Substituting \( B \) into equation (2):
\[ -4A + C = 0 \implies C = 4A \]

Substituting \( B \) and \( C \) into equation (3):
\[ 4A - 4(-A) + 2(4A) = 1 \]
\[ 4A + 4A + 8A = 1 \implies 16A = 1 \implies A = \frac{1}{16} \]

Now substituting back to find \( B \) and \( C \):
\[ B = -\frac{1}{16}, \quad C = 4 \cdot \frac{1}{16} = \frac{1}{4} \]

### Step 7: Rewrite the Integral
Now we can rewrite the integral using the values of \( A \), \( B \), and \( C \):
\[ \int \left( \frac{1/16}{x+2} - \frac{1/16}{x-2} + \frac{1/4}{(x-2)^2} \right) \, dx \]

### Step 8: Integrate Each Term
Now we integrate each term separately:
\[ = \frac{1}{16} \int \frac{1}{x+2} \, dx - \frac{1}{16} \int \frac{1}{x-2} \, dx + \frac{1}{4} \int \frac{1}{(x-2)^2} \, dx \]
\[ = \frac{1}{16} \ln |x+2| - \frac{1}{16} \ln |x-2| - \frac{1}{4(x-2)} + C \]

### Step 9: Combine the Logarithmic Terms
Combining the logarithmic terms:
\[ = \frac{1}{16} \left( \ln |x+2| - \ln |x-2| \right) - \frac{1}{4(x-2)} + C \]
\[ = \frac{1}{16} \ln \left| \frac{x+2}{x-2} \right| - \frac{1}{4(x-2)} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1}{(x-2)(x+2)(x-2)} \, dx = \frac{1}{16} \ln \left| \frac{x+2}{x-2} \right| - \frac{1}{4(x-2)} + C \]