Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

${\int }_2^3\frac{1}{x\sqrt{\ln x}}dx$

Short answer

The solution of the given integral is ${\int }_2^3\frac{1}{x\sqrt{\ln x}}dx=2\left[\sqrt{\ln 3}-\sqrt{\ln 2}\right]$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

${\int }_2^3\frac{1}{x\sqrt{\ln x}}dx$

Step 2. Using the substitution method. 

Let

$$\begin{gathered} u=\ln x \\ \frac{du}{dx}=\frac{1}{x} \\ du=\frac{1}{x}dx \end{gathered}$$

Step 3. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx={\int }_{x=2}^{x=3}\frac{1}{\sqrt{u}}du \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx={\int }_{x=2}^{x=3}\frac{1}{u^{1/2}}du \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx={\int }_{x=2}^{x=3}{u}^{-1/2}du \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx={\left[\frac{u^{-1/2+1}}{-1/2+1}\right]}_{x=2}^{x=3} \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx={\left[\frac{u^{1/2}}{1/2}\right]}_{x=2}^{x=3} \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx=2{\left[\sqrt{u}\right]}_2^3 \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx=2{\left[\sqrt{\ln x}\right]}_{x=2}^{x=3} \\ {\int }_2^3\frac{1}{x\sqrt{\ln x}}dx=2\left[\sqrt{\ln 3}-\sqrt{\ln 2}\right] \end{gathered}$$