Answer
$$\int^8_1\Big(\frac{2}{3x}-\frac{8}{x^2}\Big)dx=\ln 4-7$$
Work Step by Step
$$A=\int^8_1\Big(\frac{2}{3x}-\frac{8}{x^2}\Big)dx$$ $$A=\frac{2}{3}\int^8_1\frac{1}{x}dx+8\int^8_1-\frac{1}{x^2}dx$$ $$A=\frac{2}{3}(\ln x)\Big]^8_1+8\times\Big(\frac{1}{x}\Big)\Big]^8_1$$ $$A=\frac{2}{3}(\ln8-\ln1)+8\Big(\frac{1}{8}-1\Big)$$ $$A=\frac{2\ln8}{3}+1-8$$ $$A=\frac{2\ln8}{3}-7$$ $$A=\frac{2\ln2^3}{3}-7$$ $$A=2\ln2-7$$ $$A=\ln4-7$$
