Express (5x-2)/(x3(x+2)) in partial fraction.

The denominator has a repeated linear factor $x^3$ and a distinct linear factor $(x+2)$.

For $x^3$: $A+D = 0 \quad (1)$
For $x^2$: $2A+B = 0 \quad (2)$
For $x^1$: $2B+C = 5 \quad (3)$
For $x^0$ (constant): $2C = -2 \quad (4)$

From (4): $C = -1$.
Substitute $C=-1$ into (3): $2B + (-1) = 5 \implies 2B = 6 \implies B = 3$.
Substitute $B=3$ into (2): $2A + 3 = 0 \implies 2A = -3 \implies A = -\frac{3}{2}$.
Substitute $A=-\frac{3}{2}$ into (1): $-\frac{3}{2} + D = 0 \implies D = \frac{3}{2}$.

The final answer is $\boxed{-\frac{3}{2x} + \frac{3}{x^2} - \frac{1}{x^3} + \frac{3}{2(x+2)}}$.

2. Integrate $\int \frac{5x-2}{x^3(x+2)}\,dx$.

Recall that $\int x^n\,dx = \frac{x^{n+1}}{n+1} + K$ for $n \ne -1$, and $\int \frac{1}{x}\,dx = \ln|x| + K$.

This can be rewritten using logarithm properties $\ln a - \ln b = \ln \frac{a}{b}$:

The final answer is $\boxed{\frac{3}{2}\ln\left|\frac{x+2}{x}\right| - \frac{3}{x} + \frac{1}{2x^2} + K}$.

3. Solve $\frac{x+1}{x-1} \ge 2$.

Numerator: $-x+3 = 0 \implies x = 3$
Denominator: $x-1 = 0 \implies x = 1$
Note that $x \ne 1$ because the denominator cannot be zero.

• For $x < 1$ (e.g., $x=0$): $\frac{-0+3}{0-1} = \frac{3}{-1} = -3$. This is not $\ge 0$.
• For $1 < x \le 3$ (e.g., $x=2$): $\frac{-2+3}{2-1} = \frac{1}{1} = 1$. This is $\ge 0$. So, this interval is part of the solution.
• For $x > 3$ (e.g., $x=4$): $\frac{-4+3}{4-1} = \frac{-1}{3}$. This is not $\ge 0$.

The inequality holds for $1 < x \le 3$.
The final answer is $\boxed{1 < x \le 3}$.