To divide the polynomial \( x^3 + 3x^2 - 2x + 1 \) by \( 2x - 1 \), we can use polynomial long division. Here are the steps:

1. **Divide the leading term of the numerator by the leading term of the denominator:**
\[
\frac{x^3}{2x} = \frac{1}{2}x^2
\]
So, the first term of the quotient is \(\frac{1}{2}x^2\).

2. **Multiply the entire divisor \(2x - 1\) by \(\frac{1}{2}x^2\):**
\[
\left(\frac{1}{2}x^2\right)(2x - 1) = x^3 - \frac{1}{2}x^2
\]

3. **Subtract this result from the original polynomial:**
\[
(x^3 + 3x^2 - 2x + 1) - (x^3 - \frac{1}{2}x^2) = 3x^2 + \frac{1}{2}x^2 - 2x + 1 = \frac{7}{2}x^2 - 2x + 1
\]

4. **Repeat the process with the new polynomial \(\frac{7}{2}x^2 - 2x + 1\):**
\[
\frac{\frac{7}{2}x^2}{2x} = \frac{7}{4}x
\]
So, the next term of the quotient is \(\frac{7}{4}x\).

5. **Multiply the entire divisor \(2x - 1\) by \(\frac{7}{4}x\):**
\[
\left(\frac{7}{4}x\right)(2x - 1) = \frac{7}{2}x^2 - \frac{7}{4}x
\]

6. **Subtract this result from the current polynomial:**
\[
\left(\frac{7}{2}x^2 - 2x + 1\right) - \left(\frac{7}{2}x^2 - \frac{7}{4}x\right) = -2x + \frac{7}{4}x + 1 = -\frac{1}{4}x + 1
\]

7. **Repeat the process with the new polynomial \(-\frac{1}{4}x + 1\):**
\[
\frac{-\frac{1}{4}x}{2x} = -\frac{1}{8}
\]
So, the next term of the quotient is \(-\frac{1}{8}\).

8. **Multiply the entire divisor \(2x - 1\) by \(-\frac{1}{8}\):**
\[
\left(-\frac{1}{8}\right)(2x - 1) = -\frac{1}{4}x + \frac{1}{8}
\]

9. **Subtract this result from the current polynomial:**
\[
\left(-\frac{1}{4}x + 1\right) - \left(-\frac{1}{4}x + \frac{1}{8}\right) = 1 - \frac{1}{8} = \frac{7}{8}
\]

So, the quotient is \(\frac{1}{2}x^2 + \frac{7}{4}x - \frac{1}{8}\) and the remainder is \(\frac{7}{8}\).

Thus, the result of the division is:
\[
\frac{x^3 + 3x^2 - 2x + 1}{2x - 1} = \frac{1}{2}x^2 + \frac{7}{4}x - \frac{1}{8} + \frac{\frac{7}{8}}{2x - 1}
\]