The problem asks us to express the given rational function $\frac{1}{x^2(2x-1)}$ in partial fractions.

1. Problem Description

The problem asks us to express the given rational function $\frac{1}{x^2(2x-1)}$ in partial fractions.

2. Solution Steps

Since the denominator is $x^2(2x-1)$, we can express the given fraction as:

$$\frac{1}{x^2(2x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-1}$$

Multiplying both sides by $x^2(2x-1)$ yields:

$$1 = A(x)(2x-1) + B(2x-1) + C(x^2)$$

$$1 = A(2x^2-x) + B(2x-1) + Cx^2$$

$$1 = 2Ax^2 - Ax + 2Bx - B + Cx^2$$

$$1 = (2A+C)x^2 + (-A+2B)x - B$$

Comparing the coefficients of $x^2$, $x$ and the constant terms, we get the following system of equations:

$$2A + C = 0 \quad (1)$$

$$-A + 2B = 0 \quad (2)$$

$$-B = 1 \quad (3)$$

From equation (3), we have $B = -1$.

Substituting $B = -1$ into equation (2), we have $-A + 2(-1) = 0$, so $-A - 2 = 0$, which gives $A = -2$.

Substituting $A = -2$ into equation (1), we have $2(-2) + C = 0$, so $-4 + C = 0$, which gives $C = 4$.

Therefore, $A = -2$, $B = -1$, and $C = 4$.

Thus, we have

$$\frac{1}{x^2(2x-1)} = \frac{-2}{x} + \frac{-1}{x^2} + \frac{4}{2x-1}$$

$$\frac{1}{x^2(2x-1)} = -\frac{2}{x} - \frac{1}{x^2} + \frac{4}{2x-1}$$

3. Final Answer

The partial fraction decomposition is

$$-\frac{2}{x} - \frac{1}{x^2} + \frac{4}{2x-1}$$