A) Step 1: Express the rational function in partial fractions.

By comparing coefficients:
For the constant term: $2B = -2 \implies B = -1$.
For the coefficient of $x$: $2A+B = 5$. Substitute $B=-1$: $2A-1 = 5 \implies 2A = 6 \implies A = 3$.
For the coefficient of $x^2$: $A+C = 0$. Substitute $A=3$: $3+C = 0 \implies C = -3$.

Combine the logarithmic terms:

The partial fraction decomposition is $\boxed{\frac{3}{x} - \frac{1}{x^2} - \frac{3}{x+2}}$ and the integral is $\boxed{3 \ln\left|\frac{x}{x+2}\right| + \frac{1}{x} + K}$.

The critical points are where the numerator is zero or the denominator is zero.
Numerator: $-x+3 = 0 \implies x=3$.
Denominator: $x-1 = 0 \implies x=1$.

The critical points $x=1$ and $x=3$ divide the number line into three intervals: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

• For $x < 1$ (e.g., $x=0$): $\frac{-0+3}{0-1} = \frac{3}{-1} = -3$. This is not $\ge 0$.
• For $1 < x < 3$ (e.g., $x=2$): $\frac{-2+3}{2-1} = \frac{1}{1} = 1$. This is $\ge 0$.
• For $x > 3$ (e.g., $x=4$): $\frac{-4+3}{4-1} = \frac{-1}{3}$. This is not $\ge 0$.

• At $x=3$: $\frac{-3+3}{3-1} = \frac{0}{2} = 0$. Since $0 \ge 0$, $x=3$ is included in the solution.
• At $x=1$: The denominator is zero, so the expression is undefined. Thus, $x=1$ is not included.

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

When "ACT" is treated as a single unit, we consider it as one block.
The letters used in "ACT" are one A, one C, and one T.
Remaining letters from STATISTICS:
S: 3
T: $3-1 = 2$
A: $1-1 = 0$
I: 2
C: $1-1 = 0$

The items to arrange are the block (ACT) and the remaining letters:
(ACT): 1 unit
S: 3 times
T: 2 times
I: 2 times

The total number of items to arrange is $1+3+2+2 = 8$.

This is a permutation with repetitions. The formula is $\frac{N!}{n_1