Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 29

$$\int^{2}_{1}\frac{2\ln x}{x}dx=(\ln2)^2$$

$$A=\int^{2}_{1}\frac{2\ln x}{x}dx$$ We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$ For $x=1$, we have $$u=\ln1=0$$ For $x=2$, we have $$u=\ln2$$ Therefore, $$A=\int^{\ln2}_02udu=\frac{2u^2}{2}\Big]^{\ln2}_0=u^2\Big]^{\ln2}_0$$ $$A=(\ln2)^2-0=(\ln2)^2$$