Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 32

$$\int^{16}_{2}\frac{dx}{2x\sqrt{\ln x}}=\sqrt{\ln16}-\sqrt{\ln2}=\sqrt{\ln2}$$

$$A=\int^{16}_{2}\frac{dx}{2x\sqrt{\ln x}}$$ We set $u=\sqrt{\ln x}$, which means $$du=\frac{(\ln x)'}{2\sqrt{\ln x}}dx=\frac{dx}{2x\sqrt{\ln x}}$$ For $x=16$, we have $$u=\sqrt{\ln16}$$ For $x=2$, we have $$u=\sqrt{\ln2}$$ Therefore, $$A=\int^{\sqrt{\ln16}}_{\sqrt{\ln2}}du=u\Big]^{\sqrt{\ln16}}_{\sqrt{\ln2}}$$ $$A=\sqrt{\ln16}-\sqrt{\ln2}$$