Question

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

$\int \frac{\sqrt{\ln x}}{x}\mathrm{d}x$

Short answer

The solution of the given integral is $\int \frac{\sqrt{\ln x}}{x}\mathrm{d}x=\frac{2}{3}{\left(\mathrm{lnx}\right)}^{3/2}+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{\sqrt{\ln x}}{x}\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

Let

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

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \frac{\sqrt{\ln x}}{x}\mathrm{d}x=\int \sqrt{u}\mathrm{du} \\ \int \frac{\sqrt{\mathrm{lnx}}}{\mathrm{x}}\mathrm{dx}=\int {\mathrm{u}}^{1/2}\mathrm{du} \\ \int \frac{\sqrt{\mathrm{lnx}}}{\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{u}}^{1/2+1}}{1/2+1}+\mathrm{C} \\ \int \frac{\sqrt{\mathrm{lnx}}}{\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{u}}^{3/2}}{3/2}+\mathrm{C} \\ \int \frac{\sqrt{\mathrm{lnx}}}{\mathrm{x}}\mathrm{dx}=\frac{2}{3}{\mathrm{u}}^{3/2}+\mathrm{C} \\ \int \frac{\sqrt{\mathrm{lnx}}}{\mathrm{x}}\mathrm{dx}=\frac{2}{3}{\left(\mathrm{lnx}\right)}^{3/2}+\mathrm{C} \end{gathered}$$