Integral of $$$\frac{\ln\left(3 x\right)}{4 x}$$$

Find $$$\int \frac{\ln\left(3 x\right)}{4 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{\ln{\left(3 x \right)}}{x}$$$:

$${\color{red}{\int{\frac{\ln{\left(3 x \right)}}{4 x} d x}}} = {\color{red}{\left(\frac{\int{\frac{\ln{\left(3 x \right)}}{x} d x}}{4}\right)}}$$

Let $$$u=\ln{\left(3 x \right)}$$$.

Then $$$du=\left(\ln{\left(3 x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (steps can be seen »), and we have that $$$\frac{dx}{x} = du$$$.

Thus,

$$\frac{{\color{red}{\int{\frac{\ln{\left(3 x \right)}}{x} d x}}}}{4} = \frac{{\color{red}{\int{u d u}}}}{4}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\frac{{\color{red}{\int{u d u}}}}{4}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{4}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{4}$$

Recall that $$$u=\ln{\left(3 x \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{8} = \frac{{\color{red}{\ln{\left(3 x \right)}}}^{2}}{8}$$

Therefore,

$$\int{\frac{\ln{\left(3 x \right)}}{4 x} d x} = \frac{\ln{\left(3 x \right)}^{2}}{8}$$

Simplify:

$$\int{\frac{\ln{\left(3 x \right)}}{4 x} d x} = \frac{\left(\ln{\left(x \right)} + \ln{\left(3 \right)}\right)^{2}}{8}$$

Add the constant of integration:

$$\int{\frac{\ln{\left(3 x \right)}}{4 x} d x} = \frac{\left(\ln{\left(x \right)} + \ln{\left(3 \right)}\right)^{2}}{8}+C$$

$$$\int \frac{\ln\left(3 x\right)}{4 x}\, dx = \frac{\left(\ln\left(x\right) + \ln\left(3\right)\right)^{2}}{8} + C$$$A