Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 19

You need to use partial fraction decomposition, such that:
(x^2+1)/((x-3)(x-2)^2) = a/(x-3) + b/(x-2) + c/((x-2)^2)
x^2+1 = a(x-2)^2 + b(x-2)(x-3) + c(x-3)
x^2+1 = ax^2 - 4ax + 4a + bx^2 - 5bx + 6b + cx - 3c
Group the terms:
x^2+1 = x^2(a+b) + x(-4a-5b+c) + 4a + 6b - 3c
a + b = 1 => a = 1-b
-4a-5b+c = 0 => c = 4a+5b
6b - 3c = 1 => 6b - 3(4a+5b) = 1 => -9b - 12a = 1
-9b - 12(1-b) = 1 => 3b - 12 = 1 => 3b = 13 => b = 13/3
a = -10/3
c = -40/3 + 65/3 => c = 25/3
(x^2+1)/((x-3)(x-2)^2) = -10/(3(x-3)) + 13/(3(x-2)) + 25/(3(x-2)^2)
Taking integral both sides yields:
int (x^2+1)/((x-3)(x-2)^2) dx = int -10/(3(x-3)) dx + int 13/(3(x-2)) dx + int 25/(3(x-2)^2)dx
int (x^2+1)/((x-3)(x-2)^2) dx = -10/3*ln|x-3| + 13/3ln|x-2| - 25/(3(x-2)) + c
Hence, evaluating the give integral yields int (x^2+1)/((x-3)(x-2)^2) dx = -10/3*ln|x-3| + 13/3ln|x-2| - 25/(3(x-2)) + c.