Answer
$$\int^{2}_1(\frac{1}{x^2}-\frac{4}{x^3})dx=-1$$
Work Step by Step
$$A=\int^{2}_1(\frac{1}{x^2}-\frac{4}{x^3})dx$$ $$A=\int^{2}_1(x^{-2}-4x^{-3})dx$$ According to Table 1, we have $$\int(x^n)dx=\frac{x^{n+1}}{n+1}+C (n\ne-1)$$ $$\int cf(x)dx=c\int f(x)$$ Therefore, $$A=(\frac{x^{-1}}{-1}-4\frac{x^{-2}}{-2})\Bigg]^2_1$$ $$A=(\frac{-1}{x}+\frac{2}{x^2})\Bigg]^2_1$$ $$A=(\frac{-1}{2}+\frac{2}{2^2})-(\frac{-1}{1}+\frac{2}{1^2})$$ $$A=(\frac{-1}{2}+\frac{1}{2})-(-1+2)$$ $$A=-1$$
