Question

evaluate the integral
\\( \int \frac{\ln \sqrt{8x}}{x} dx \\)

\\( \int \frac{\ln \sqrt{8x}}{x} dx = \square \\)
(use parentheses to clearly denote the argument of each function )

Explanation:

Step1: Simplify the logarithm

First, use the property of logarithms \(\ln\sqrt{a}=\frac{1}{2}\ln a\). So, \(\ln\sqrt{8x}=\frac{1}{2}\ln(8x)\). Then, the integral becomes \(\int\frac{\frac{1}{2}\ln(8x)}{x}dx=\frac{1}{2}\int\frac{\ln(8x)}{x}dx\).

Step2: Use substitution

Let \(u = \ln(8x)\). Then, find the derivative of \(u\) with respect to \(x\). Using the chain rule, \(\frac{du}{dx}=\frac{1}{8x}\cdot8=\frac{1}{x}\), so \(du=\frac{1}{x}dx\).

Step3: Substitute into the integral

Substitute \(u\) and \(du\) into the integral. The integral \(\frac{1}{2}\int\frac{\ln(8x)}{x}dx\) becomes \(\frac{1}{2}\int u\ du\).

Step4: Integrate with respect to \(u\)

The integral of \(u\) with respect to \(u\) is \(\frac{1}{2}u^{2}+C\) (where \(C\) is the constant of integration). So, \(\frac{1}{2}\int u\ du=\frac{1}{2}\cdot\frac{1}{2}u^{2}+C=\frac{1}{4}u^{2}+C\).

Step5: Substitute back \(u\)

Substitute back \(u = \ln(8x)\) into the expression. We get \(\frac{1}{4}(\ln(8x))^{2}+C\).

Answer:

\(\frac{1}{4}(\ln(8x))^{2}+C\)