Answer
$$\int^4_1\Big(\frac{x}{8}+\frac{1}{2x}\Big)=\ln2+\frac{15}{16}$$
Work Step by Step
$$A=\int^4_1\Big(\frac{x}{8}+\frac{1}{2x}\Big)dx$$ $$A=\frac{1}{8}\int^4_1xdx+\frac{1}{2}\int^4_1\frac{1}{x}dx$$ $$A=\frac{1}{8}\times\frac{x^2}{2}\Big]^4_1+\frac{1}{2}(\ln|x|)\Big]^4_1$$ $$A=\frac{x^2}{16}\Big]^4_1+\frac{1}{2}(\ln4-\ln1)$$ $$A=\frac{16}{16}-\frac{1}{16}+\frac{\ln4-0}{2}$$ $$A=\frac{\ln4}{2}+\frac{15}{16}$$ $$A=\ln2+\frac{15}{16}$$
