Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 38

You need to use partial fraction decomposition, such that:
(x^3+2x^2+3x-2)/((x^2+2x+2)^2) = (ax+b)/(x^2+2x+2) + (cx+d)/((x^2+2x+2)^2)
x^3+2x^2+3x-2 = (ax+b)(x^2+2x+2) + cx + d
x^3+2x^2+3x-2 = ax^3 + 2ax^2 + 2ax + bx^2 + 2bx + 2b + cx + d
x^3+2x^2+3x-2 = ax^3 + x^2(2a+b) + x(2a+2b+c) + 2b+d
a = 1
2a+b = 2 => 2 + b = 2 => b = 0
2a+2b+c = 3 => 2 + 0 + c = 3 => c = 1
2b+d = -2 => d = -2
(x^3+2x^2+3x-2)/((x^2+2x+2)^2) = 1/(x^2+2x+2) + (x-2)/((x^2+2x+2)^2)
Integrating both sides yields:
int (x^3+2x^2+3x-2)/((x^2+2x+2)^2) dx = int 1/(x^2+2x+2) dx + int (x-2)/((x^2+2x+2)^2) dx
int 1/(x^2+2x+2) dx = int 1/((x+1)^2+1) dx = arctan (x+1) + c
int (x-2)/((x^2+2x+2)^2) dx = (1/2)int (2x-4)/((x^2+2x+2)^2) dx
(1/2)int (2x + 2 - 2 -4)/((x^2+2x+2)^2) dx = (1/2)int (2x + 2)/((x^2+2x+2)^2) dx - 3int 1/(((x+1)^2+1)^2) dx
You need to solve (1/2)int (2x + 2)/((x^2+2x+2)^2) dx using substitution x^2+2x+2 = t => (2x+2)dx = dt
(1/2)int (2x + 2)/((x^2+2x+2)^2) dx = (1/2) int dt/(t^2) = -1/(2t) = -1/(2(x^2+2x+2)) + c
3int 1/(((x+1)^2+1)^2) dx = (3/2)(arctan(x+1)) + (3/2)(x+1)/((x+1)^2+1) + c
Hence, evaluating the integral yields int (x^3+2x^2+3x-2)/((x^2+2x+2)^2) dx = arctan (x+1) - 1/(2(x^2+2x+2)) - (3/2)(arctan(x+1)) - (3/2)(x+1)/((x+1)^2+1) + c