Answer: (a) \(\dfrac{1-2\ln3x}{x^3}\); (b) \(-\dfrac{1}{4x^2}-\dfrac{\ln3x}{2x^2}+C\).

We start with the main method. Rewrite the expression into standard integrable terms before applying the usual integration rules.

Write

\(y=x^{-2}\ln3x.\)

Using the product rule,

\(\frac{dy}{dx}=x^{-2}\cdot\frac1x+\ln3x\cdot(-2x^{-3}).\)

So

\(\frac{dy}{dx}=\frac{1}{x^3}-\frac{2\ln3x}{x^3}.\)

Hence

\(\frac{dy}{dx}=\frac{1-2\ln3x}{x^3}.\)

For part (b), rearrange the result from part (a):

\(\frac{dy}{dx}=\frac1{x^3}-\frac{2\ln3x}{x^3}.\)

Therefore

\(\frac{2\ln3x}{x^3}=\frac1{x^3}-\frac{dy}{dx}.\)

So

\(\int\frac{\ln3x}{x^3}\,dx =\frac12\int x^{-3}\,dx-\frac12y.\)

Now

\(\int x^{-3}\,dx=-\frac{1}{2x^2}.\)

Also \(y=\frac{\ln3x}{x^2}\).

Hence

\(\int\frac{\ln3x}{x^3}\,dx =-\frac{1}{4x^2}-\frac{\ln3x}{2x^2}+C.\)

This completes the solution and gives the required result.