Given,

\(\int \ln(\frac{x}{3})dx\)

a).

By using the table of integrals, we have \(\int \ln x dx = x\ln (x)-x+C\)

Therefore,

\(\int \ln (\frac{x}{3})dx=\frac{\frac{x}{3}\times\ln(\frac{x}{3})-\frac{x}{3}}{\frac{1}{3}}+C\) (x-coefficient must be divided)

\(=x\ln(\frac{x}{3})-x+C\)

Step 2

b).

\(\int \ln (\frac{x}{3})dx=\int \ln (\frac{x}{3})\times1dx\)

Now integrating by parts, we get

\(\int\ln(\frac{x}{3})\times1 dx=\ln(\frac{x}{3})\times\int 1 dx - \int [\frac{d}{dx}[\ln(\frac{x}{3})]\times\int 1 dx]+C\)

\(=\ln(\frac{x}{3})\times x - \int(\frac{1}{\frac{x}{3}}\times\frac{1}{3}\times x)dx+C\)

\(=\ln(\frac{x}{3})\times x-\int 1 dx+C\)

\(=x\ln(\frac{x}{3})-x+C\)