Calculus 8th Edition

$2$
We can rewrite the terms in the integrand to remove the fractions, which can be easier to integrate. This would make our new integrand is $\int_1^2 (x^{-2} - 4x^{-3}dx)$ We can treat the $4$ as a constant and can multiply it by the integral of $x^{-3}$. Our integral would be $-x^{-1} - (4) \frac{(-x^{-2})}{{-2}} \Big|_1^2$ Before substituting our bounds of integration, we can simplify the expression as $-\frac{1}{x} - (4)(\frac{1}{2x^2})$ $= -\frac{1}{x} - \frac{2}{x^2}$ $= -\frac{x+2}{x^2}$ We now substitute $2$ and $1$ into $x$ and get $-\frac{(2) + 2}{(2)^2} - (- \frac {(1) + 2}{(1)^2})$ $= -\frac{4}{4} + \frac{3}{1}$ $= -1 + 3 = 2$