`int_1^2 (1/x^2 - 4/x^3) dx` Evaluate the integral

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Chapter 5, 5.4 - Problem 28 - 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)`

`int_1^2 (1/x^2 - 4/x^3)dx = int_1^2 1/x^2 dx - int_1^2 4/x^3 dx`

You need to use the formula `int 1/x^n dx = (x^(-n+1))/(1-n) + c:`

`int_1^2 (1/x^2 - 4/x^3)dx = (-1/x + 4/(2x^2))|_1^2`

`int_1^2 (1/x^2 - 4/x^3)dx = -1/2 + 2/4 + 1 - 2`

`int_1^2 (1/x^2 - 4/x^3)dx = -1`

Hence, evaluating the definite integral yields `int_1^2 (1/x^2 - 4/x^3)dx = -1.`

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