Integral of $$$\ln\left(6 x\right)$$$

Let $$$u=6 x$$$.

Then $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{6}$$$.

Therefore,

$${\color{red}{\int{\ln{\left(6 x \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{6} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{6} d u}}} = {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{6}\right)}}$$

For the integral $$$\int{\ln{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{\theta} \operatorname{dv} = \operatorname{\theta}\operatorname{v} - \int \operatorname{v} \operatorname{d\theta}$$$.

Let $$$\operatorname{\theta}=\ln{\left(u \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{d\theta}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

Therefore,

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{6}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{6}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{6}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$\frac{u \ln{\left(u \right)}}{6} - \frac{{\color{red}{\int{1 d u}}}}{6} = \frac{u \ln{\left(u \right)}}{6} - \frac{{\color{red}{u}}}{6}$$

Recall that $$$u=6 x$$$:

$$- \frac{{\color{red}{u}}}{6} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{6} = - \frac{{\color{red}{\left(6 x\right)}}}{6} + \frac{{\color{red}{\left(6 x\right)}} \ln{\left({\color{red}{\left(6 x\right)}} \right)}}{6}$$

Therefore,

$$\int{\ln{\left(6 x \right)} d x} = x \ln{\left(6 x \right)} - x$$

Simplify:

$$\int{\ln{\left(6 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(6 \right)}\right)$$

Add the constant of integration:

$$\int{\ln{\left(6 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(6 \right)}\right)+C$$

$$$\int \ln\left(6 x\right)\, dx = x \left(\ln\left(x\right) - 1 + \ln\left(6\right)\right) + C$$$A