`int_1^2 (x-1)^3/x^2 dx` Evaluate the integral

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Chapter 5, 5.4 - Problem 42 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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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.`

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