(int\frac{ln^{2}3x}{x}dxquadint\frac{\tan(ln x)}{x}dx)

Explanation:

Step1: First - integral substitution

Let (u = \ln x), then (du=\frac{1}{x}dx). For the integral (\int\frac{\ln^{2}3x}{x}dx), since (\ln(3x)=\ln3+\ln x), the integral becomes (\int(\ln3 + u)^{2}du=\int(\ln^{2}3 + 2\ln3u+u^{2})du).

Step2: Integrate term - by - term

(\int(\ln^{2}3 + 2\ln3u+u^{2})du=\int\ln^{2}3du+2\ln3\int udu+\int u^{2}du). We know that (\int\ln^{2}3du=\ln^{2}3\int du=\ln^{2}3u), (2\ln3\int udu = 2\ln3\times\frac{u^{2}}{2}=\ln3u^{2}), and (\int u^{2}du=\frac{u^{3}}{3}). Substituting back (u = \ln x), we get (\ln^{2}3\ln x+\ln3\ln^{2}x+\frac{\ln^{3}x}{3}+C). For the integral (\int\frac{\tan(\ln x)}{x}dx), with (u = \ln x) and (du=\frac{1}{x}dx), it becomes (\int\tan udu).

Step3: Integrate (\int\tan udu)

We know that (\int\tan udu=-\ln|\cos u|+C). Substituting back (u = \ln x), we get (-\ln|\cos(\ln x)|+C).

Answer:

(\int\frac{\ln^{2}3x}{x}dx=\ln^{2}3\ln x+\ln3\ln^{2}x+\frac{\ln^{3}x}{3}+C); (\int\frac{\tan(\ln x)}{x}dx=-\ln|\cos(\ln x)|+C)