`I = int( 6 x^(3) + x^(2)-2x + 1) /( 2x - 1) dx`.

Answer

To solve the integral \( I = \int \frac{6x^3 + x^2 - 2x + 1}{2x - 1} \, dx \), we will follow these steps:

### Step 1: Polynomial Long Division
Since the degree of the numerator (3) is greater than the degree of the denominator (1), we will perform polynomial long division.

1. Divide \( 6x^3 \) by \( 2x \) to get \( 3x^2 \).
2. Multiply \( 3x^2 \) by \( 2x - 1 \) to get \( 6x^3 - 3x^2 \).
3. Subtract this from the original polynomial:
\[ (6x^3 + x^2 - 2x + 1) - (6x^3 - 3x^2) = 4x^2 - 2x + 1 \]

### Step 2: Continue Polynomial Long Division
Now, we divide \( 4x^2 \) by \( 2x \) to get \( 2x \).

1. Multiply \( 2x \) by \( 2x - 1 \) to get \( 4x^2 - 2x \).
2. Subtract this from the current polynomial:
\[ (4x^2 - 2x + 1) - (4x^2 - 2x) = 1 \]

### Step 3: Write the Integral
Now we can express the integral as:
\[ I = \int (3x^2 + 2x + \frac{1}{2x - 1}) \, dx \]

### Step 4: Integrate Each Term
Now we will integrate each term separately:

1. \( \int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3 \)
2. \( \int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2 \)
3. For the last term, \( \int \frac{1}{2x - 1} \, dx \):
- This can be integrated as \( \frac{1}{2} \ln |2x - 1| \).

### Step 5: Combine Results
Putting it all together, we have:
\[ I = x^3 + x^2 + \frac{1}{2} \ln |2x - 1| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = x^3 + x^2 + \frac{1}{2} \ln |2x - 1| + C \]