Answer
$\frac{19ln2}{2} \approx 6.58$
Work Step by Step
$= \int_1^4 \int_1^2 (\frac{x}{y} + \frac{y}{x}) dy dx$
$= \int_1^4 (xlny + \frac{y^2}{2x})\big|_1^2 dx$
$= \int_1^4 [(xln2 + \frac{2}{x}) – (0 + \frac{1}{2x})] dx$
$= \int_1^4 (xln2 + \frac{2}{x} – \frac{1}{2x}) dx$
$= (\frac{x^2 ln2}{2} + 2lnx - \frac{lnx}{2})\big|_1^4$
$= [(8ln2 + 2ln4 - \frac{ln4}{2}) – (\frac{ln2}{2} + 0 - 0)]$
$= 8ln2 + 4ln2 – ln2 – \frac{ln2}{2}$
$= \frac{19ln2}{2}$
$\approx 6.58$
