Answer
$\frac{21}{2}\ln2$
Work Step by Step
We begin with the iterated integral:
$$\int_1^4\int_1^2\bigg(\frac{x}{y}+\frac{y}{x}\bigg)\,dy\,dx$$
Solving the inner integral, we get:
$$\int_1^4\bigg[x\ln{|y|}+\frac{y^2}{2x}\bigg]_{y=1}^{y=2}\,dx\\
=\int_1^4\bigg(x\ln{2}+\frac{4}{2x}-0-\frac{1}{2x}\bigg)\,dx\\
=\int_1^4\bigg(x\ln{2}+\frac{3}{2x}\bigg)\,dx$$
Integrating, we get:
$$\bigg[\frac{x^2\ln{2}}{2}+\frac{3}{2}\ln{|x|}\bigg]_{1}^4=\\
8\ln2+\frac{3}{2}\ln4-\frac{1}{2}\ln2-0\\
=\frac{15}{2}\ln{2}+\frac{3}{2}ln{4}\\
=\frac{15}{2}\ln{2}+\frac{3}ln{2}\\
=\frac{21}{2}\ln2$$
