a) Step 1: Express the rational function in partial fractions. We set up the partial fraction decomposition for $\frac{5x-2}{x^2(x+2)}$: $$ \frac{5x-2}{x^2(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2} $$ Multiply both sides by $x^2(x+2)$ to clear the denominators: $$ 5x-2 = A x(x+2) + B(x+2) + C x^2 $$ $$ 5x-2 = Ax^2 + 2Ax + Bx + 2B + C x^2 $$ $$ 5x-2 = (A+C)x^2 + (2A+B)x + 2B $$ Step 2: Solve for the constants A, B, and C. 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 $A=3$: $3+C = 0 \implies C = -3$. Step 3: Write the partial fraction decomposition. $$ \frac{5x-2}{x^2(x+2)} = \frac{3}{x} - \frac{1}{x^2} - \frac{3}{x+2} $$ Step 4: Integrate the partial fractions. $$ \int \frac{5x-2}{x^2(x+2)}dx = \int \left(\frac{3}{x} - \frac{1}{x^2} - \frac{3}{x+2}\right)dx $$ $$ = \int \frac{3}{x}dx - \int x^{-2}dx - \int \frac{3}{x+2}dx $$ $$ = 3 \ln|x| - \frac{x^{-1}}{-1} - 3 \ln|x+2| + K $$ $$ = 3 \ln|x| + \frac{1}{x} - 3 \ln|x+2| + K $$ Combine the logarithmic terms: $$ = 3 (\ln|x| - \ln|x+2|) + \frac{1}{x} + K $$ $$ = 3 \ln\left|\frac{x}{x+2}\right| + \frac{1}{x} + K $$ The partial fraction decomposition is $\boxed{\frac{3}{x} - \frac{1}{x^2} - \frac{3}{x+2}}$ and the integral is $\boxed{3 \ln\left|\frac{x}{x+2}\right| + \frac{1}{x} + K}$. b) Step 1: Move all terms to one side and combine into a single fraction. $$ \frac{x+1}{x-1} \ge 2 $$ $$ \frac{x+1}{x-1} - 2 \ge 0 $$ $$ \frac{x+1 - 2(x-1)}{x-1} \ge 0 $$ $$ \frac{x+1 - 2x + 2}{x-1} \ge 0 $$ $$ \frac{-x+3}{x-1} \ge 0 $$ Step 2: Find the critical points. The critical points are where the numerator is zero or the denominator is zero. Numerator: $-x+3 = 0 \implies x=3$. Denominator: $x-1 = 0 \implies x=1$. Step 3: Test intervals defined by the critical points. 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 $\ge 0$. Step 4: Check the endpoints. 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. Step 5: State the solution. The inequality holds for $1 < x \le 3$. The solution is $\boxed{1 < x \le 3}$. c) Step 1: Identify the letters and their frequencies in the word STATISTICS. The word STATISTICS has 10 letters: S: 3 T: 3 A: 1 I: 2 C: 1 Step 2: Treat the sub-word ACT as a single unit. 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: $1-1 = 0$ Step 3: List the items to be arranged and their frequencies. 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 $1+3+2+2 = 8$. Step 4: Calculate the number of distinct arrangements. This is a permutation with repetitions. The formula is $\frac{N!}{n_1