$int\frac{ln^{2}3x}{x}dx$

Explanation:

Step1: Set substitution

Let $u = \ln(3x)$, then $du=\frac{1}{x}dx$.

Step2: Rewrite integral

The integral $\int\frac{\ln^{2}(3x)}{x}dx$ becomes $\int u^{2}du$.

Step3: Integrate $u^{2}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int u^{2}du=\frac{u^{3}}{3}+C$.

Step4: Substitute back

Substitute $u=\ln(3x)$ back into the result, we get $\frac{\ln^{3}(3x)}{3}+C$.

Answer:

$\frac{\ln^{3}(3x)}{3}+C$