Answer
$1$
Work Step by Step
Consider $f(x)=\int_1^2(x-\dfrac{1}{x^2}) dx$
Use formula such as: $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
Then $f(x)=[\dfrac{x^{(1+1)}}{(1+1)}-\dfrac{x^{(-2+1)}}{(-2+1)}]_1^2$
and $[\dfrac{x^2}{2}+\dfrac{1}{x}]_1^2=2+\dfrac{1}{2}-\dfrac{1}{2}-1$
Thus, $f(x)=\int_1^2(x-\dfrac{1}{x^2}) dx=1$
