Calculus: Early Transcendentals, Chapter 5, 5.4, Section 5.4, Problem 42

You need to evaluate the definite integral using the fundamental theorem of calculus, such that: int_a^b f(x)dx = F(b) - F(a)
You need to expand the cube such that:
(x-1)^3 = x^3 - 1 - 3x(x-1)
Replacing under the integral the expansion of the cube, yields:
int_1^2 (x-1)^3/(x^2)dx = int_1^2 (x^3 - 1 - 3x(x-1))/(x^2)dx
int_1^2 (x-1)^3/(x^2)dx = int_1^2 (x^3)/(x^2)dx - int_1^2 1/(x^2)dx - int_1^2 (3x^2)/(x^2) dx + int_1^2(3x)/(x^2)dx
Reducing like terms yields:
int_1^2 (x-1)^3/(x^2)dx = int_1^2 x dx - int_1^2 x^(-2)dx - int_1^2 3dx + int_1^2 3/x dx
int_1^2 (x-1)^3/(x^2)dx = (x^2/2 + 1/x - 3x + 3ln x)|_1^2
int_1^2 (x-1)^3/(x^2)dx = (2^2/2 + 1/2 - 6 + 3ln 2 - 1/2 - 1 + 3 - 3ln 1)
int_1^2 (x-1)^3/(x^2)dx = 3log(2) - 2
Hence, evaluating the definite integral yields int_1^2 (x-1)^3/(x^2)dx = 3log(2) - 2.