Question

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

$\int \frac{\ln \sqrt{x}}{x}dx$

Short answer

The solution of the given integral is $\int \frac{\ln \sqrt{x}}{x}dx=\frac{{\left[\ln \sqrt{x}\right]}^2}{4}+C$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{\ln \sqrt{x}}{x}dx$

Step 2. Solving the given integral using substitution method. 

Let

$$\begin{gathered} u=\ln \sqrt{x} \\ \frac{du}{dx}=\frac{1}{\sqrt{x}}·\frac{1}{2\sqrt{x}} \\ \frac{du}{dx}=\frac{1}{2x} \\ \frac{1}{2}du=\frac{1}{x}dx \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \frac{\ln \sqrt{x}}{x}\mathrm{d}x=\frac{1}{2}\int u\mathrm{d}u \\ \int \frac{\ln \sqrt{x}}{x}\mathrm{d}x=\frac{1}{2}\left[\frac{u^{1+1}}{1+1}\right]+C \\ \int \frac{\ln \sqrt{x}}{x}\mathrm{d}x=\frac{1}{2}\left[\frac{u^2}{2}\right]+C \\ \int \frac{\ln \sqrt{x}}{x}\mathrm{d}x=\frac{u^2}{4}+C \\ \int \frac{\ln \sqrt{x}}{x}\mathrm{d}x=\frac{{\left[\ln \sqrt{x}\right]}^2}{4}+C \end{gathered}$$