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

Answer

To solve the integral \( I = \int \frac{x}{(x-2)(x+1)} \, dx \), we will use the method of partial fractions.

### Step-by-Step Solution:

1. **Set Up the Partial Fraction Decomposition**:
We want to express the integrand in the form:
\[ \frac{x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \]
where \( A \) and \( B \) are constants to be determined.

2. **Combine the Right Side**:
To combine the right side, we find a common denominator:
\[ \frac{A}{x-2} + \frac{B}{x+1} = \frac{A(x+1) + B(x-2)}{(x-2)(x+1)} \]
Therefore, we have:
\[ x = A(x+1) + B(x-2) \]

3. **Expand and Collect Terms**:
Expanding the right side gives:
\[ x = Ax + A + Bx - 2B \]
Combining like terms:
\[ x = (A + B)x + (A - 2B) \]

4. **Set Up the System of Equations**:
Now, we can equate coefficients from both sides:
- Coefficient of \( x \): \( A + B = 1 \)
- Constant term: \( A - 2B = 0 \)

5. **Solve the System of Equations**:
From the second equation, we can express \( A \) in terms of \( B \):
\[ A = 2B \]
Substituting this into the first equation:
\[ 2B + B = 1 \implies 3B = 1 \implies B = \frac{1}{3} \]
Now substituting back to find \( A \):
\[ A = 2B = 2 \times \frac{1}{3} = \frac{2}{3} \]

6. **Rewrite the Integral**:
Now we can rewrite the integral:
\[ I = \int \left( \frac{2/3}{x-2} + \frac{1/3}{x+1} \right) \, dx \]

7. **Integrate Each Term**:
Now we can integrate each term separately:
\[ I = \frac{2}{3} \int \frac{1}{x-2} \, dx + \frac{1}{3} \int \frac{1}{x+1} \, dx \]
The integrals are:
\[ \int \frac{1}{x-2} \, dx = \ln |x-2| + C_1 \]
\[ \int \frac{1}{x+1} \, dx = \ln |x+1| + C_2 \]

8. **Combine the Results**:
Thus, we have:
\[ I = \frac{2}{3} \ln |x-2| + \frac{1}{3} \ln |x+1| + C \]
where \( C \) is the constant of integration.

### Final Result:
\[ I = \frac{2}{3} \ln |x-2| + \frac{1}{3} \ln |x+1| + C \]