Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 36

$[\frac{(ln2)^2}{2}]\approx0.2402$

$\int_{1}^{2} (\frac{lnx}{x}) dx=\int_{1}^{2} {lnx}\cdot \frac{dx}{x}=[\frac{(lnx)^2}{2}]_{1}^{2}=[\frac{(ln2)^2}{2}]-[\frac{(ln1)^2}{2}]=[\frac{(ln2)^2}{2}]-0=[\frac{(ln2)^2}{2}]\approx0.2402$