Integral of $\ln\left(3 x\right)$

Let $u=3 x$.

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

Thus,

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

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

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

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,

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{3}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{3}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{3}$$

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

$$\frac{u \ln{\left(u \right)}}{3} - \frac{{\color{red}{\int{1 d u}}}}{3} = \frac{u \ln{\left(u \right)}}{3} - \frac{{\color{red}{u}}}{3}$$

Recall that $u=3 x$:

$$- \frac{{\color{red}{u}}}{3} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{3} = - \frac{{\color{red}{\left(3 x\right)}}}{3} + \frac{{\color{red}{\left(3 x\right)}} \ln{\left({\color{red}{\left(3 x\right)}} \right)}}{3}$$

Therefore,

$$\int{\ln{\left(3 x \right)} d x} = x \ln{\left(3 x \right)} - x$$

Simplify:

$$\int{\ln{\left(3 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(3 \right)}\right)$$

Add the constant of integration:

$$\int{\ln{\left(3 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(3 \right)}\right)+C$$

$\int \ln\left(3 x\right)\, dx = x \left(\ln\left(x\right) - 1 + \ln\left(3\right)\right) + C$A