Answer
$\displaystyle \frac{21\ln 2}{2}$
Work Step by Step
$\displaystyle \int_{1}^{4}\int_{1}^{2}(\frac{x}{y}+\frac{y}{x})dydx=\int_{1}^{4}[\int_{1}^{2}(\frac{x}{y}+\frac{y}{x})dy]dx$
treat x as a constant in the inner integral.
$=\displaystyle \int_{1}^{4}[\int_{1}^{2}(x\cdot\frac{1}{y}+\frac{1}{x}\cdot y)dy]dx$
$=\displaystyle \int_{1}^{4}[x\ln|y|+\frac{1}{x}\cdot\frac{y^{2}}{2}]_{y=1}^{y=2}dx$
$=\displaystyle \int_{1}^{4}[x\ln 2+\frac{1}{x}\cdot\frac{2^{2}}{2}-(x\ln 1+\frac{1}{x}\cdot\frac{1}{2}) ]dx$
$=\displaystyle \int_{1}^{4}(x\ln 2+\frac{3}{2x})dx$
$=[\displaystyle \frac{x^{2}}{2}\ln 2+\frac{3}{2}\ln|x|]_{1}^{4}$
$=\displaystyle \frac{4^{2}}{2}\ln 2+\frac{3}{2} \ln 4- (\displaystyle \frac{1^{2}}{2}\ln 2-\frac{3}{2}\ln 1)$
$=\displaystyle \frac{15}{2}\ln 2+\frac{3}{2}\ln 2^{2}$
$=\displaystyle \frac{15}{2}\ln 2+\frac{3}{2}\cdot 2\ln 2$
$=\displaystyle \frac{21\ln 2}{2}$
