Question

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

Explanation:

Step1: Use substitution for first integral

Let $u = \ln x$, then $du=\frac{1}{x}dx$. The integral $\int\frac{\ln^{2}3x}{x}dx$ becomes $\int\ln^{2}(3e^{u})du$. Since $\ln(3e^{u})=\ln 3 + u$, the integral is $\int(\ln 3 + u)^{2}du=\int(\ln^{2}3 + 2u\ln 3+u^{2})du$.
\[

$$\begin{align*} \int(\ln^{2}3 + 2u\ln 3+u^{2})du&=\int\ln^{2}3du+2\ln 3\int udu+\int u^{2}du\\ &=u\ln^{2}3 + 2\ln 3\times\frac{u^{2}}{2}+\frac{u^{3}}{3}+C_1\\ &=\ln x\ln^{2}3+\ln 3\ln^{2}x+\frac{\ln^{3}x}{3}+C_1 \end{align*}$$

\]

Step2: Use substitution for second integral

Let $t=\ln x$, then $dt = \frac{1}{x}dx$. The integral $\int\frac{\tan(\ln x)}{x}dx$ becomes $\int\tan tdt$.
We know that $\int\tan tdt=-\ln|\cos t|+C_2 = -\ln|\cos(\ln x)|+C_2$

Answer:

$\int\frac{\ln^{2}3x}{x}dx=\ln x\ln^{2}3+\ln 3\ln^{2}x+\frac{\ln^{3}x}{3}+C_1$, $\int\frac{\tan(\ln x)}{x}dx=-\ln|\cos(\ln x)|+C_2$