Question 5 Express \( \frac{25}{(x-2)(x+3)^{2}} \) in the form \( \frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^{2}} \) and hence find \( \int \frac{25}{(x-2)(x+3)^{2}} d x \)

To express \( \frac{25}{(x-2)(x+3)^{2}} \) in the form \( \frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^{2}} \), we start by setting up the equation: \[ \frac{25}{(x-2)(x+3)^{2}} = \frac{A}{x-2} + \frac{B}{x+3} + \frac{C}{(x+3)^{2}}. \] To combine the right side, we find a common denominator, which is \((x-2)(x+3)^{2}\): \[ \frac{A(x+3)^{2} + B(x-2)(x+3) + C(x-2)}{(x-2)(x+3)^{2}}. \] Now, we have: \[ 25 = A(x+3)^{2} + B(x-2)(x+3) + C(x-2). \] Next, we want to determine \(A\), \(B\), and \(C\). We can use convenient values of \(x\) to solve for these coefficients. 1. Set \(x = 2\): \[ 25 = A(2+3)^{2} \implies 25 = A(5)^{2} \implies 25 = 25A \implies A = 1. \] 2. Set \(x = -3\): \[ 25 = B(-3-2)(-3+3) + C(-3-2) \implies 25 = -5C \implies C = -5. \] 3. To find \(B\), substitute \(A\) and \(C\) back into the original equation. Let's use \(x = 0\): \[ 25 = 1(0+3)^{2} + B(0-2)(0+3) - 5(0-2). \] This gives: \[ 25 = 1(9) + B(-2)(3) + 10 \implies 25 = 9 - 6B + 10 \implies 25 = 19 - 6B \implies 6B = -6 \implies B = -1. \] Now we have: \[ A = 1, \quad B = -1, \quad C = -5. \] Thus, we can express \( \frac{25}{(x-2)(x+3)^{2}} \) as: \[ \frac{1}{x-2} - \frac{1}{x+3} - \frac{5}{(x+3)^{2}}. \] Next, we integrate: \[ \int \frac{25}{(x-2)(x+3)^{2}} \, dx = \int \left( \frac{1}{x-2} - \frac{1}{x+3} - \frac{5}{(x+3)^{2}} \right) \, dx. \] This results in: \[ \int \frac{1}{x-2} \, dx - \int \frac{1}{x+3} \, dx - 5 \int \frac{1}{(x+3)^{2}} \, dx. \] Calculating each part: 1. \( \int \frac{1}{x-2} \, dx = \ln |x-2| + C_1 \), 2. \( \int \frac{1}{x+3} \, dx = \ln |x+3| + C_2 \), 3. \( \int \frac{1}{(x+3)^{2}} \, dx = -\frac{1}{x+3} + C_3 \). Putting it all together, the final result of the integral is: \[ \ln |x-2| - \ln |x+3| + \frac{5}{x+3} + C. \] Thus, the integral of \( \frac{25}{(x-2)(x+3)^2} \) is: \[ \ln \left| \frac{x-2}{x+3} \right| + \frac{5}{x+3} + C. \]