Integral of $$$\ln\left(4 x\right)$$$

Let $$$u=4 x$$$.

Then $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{4}$$$.

Therefore,

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

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

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

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

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

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

The integral becomes

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{4}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{4}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{4}$$

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

$$\frac{u \ln{\left(u \right)}}{4} - \frac{{\color{red}{\int{1 d u}}}}{4} = \frac{u \ln{\left(u \right)}}{4} - \frac{{\color{red}{u}}}{4}$$

Recall that $$$u=4 x$$$:

$$- \frac{{\color{red}{u}}}{4} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{4} = - \frac{{\color{red}{\left(4 x\right)}}}{4} + \frac{{\color{red}{\left(4 x\right)}} \ln{\left({\color{red}{\left(4 x\right)}} \right)}}{4}$$

Therefore,

$$\int{\ln{\left(4 x \right)} d x} = x \ln{\left(4 x \right)} - x$$

Simplify:

$$\int{\ln{\left(4 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + 2 \ln{\left(2 \right)}\right)$$

Add the constant of integration:

$$\int{\ln{\left(4 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + 2 \ln{\left(2 \right)}\right)+C$$

Answer: $$$\int{\ln{\left(4 x \right)} d x}=x \left(\ln{\left(x \right)} - 1 + 2 \ln{\left(2 \right)}\right)+C$$$