Integral of $$$2 \ln\left(2 x\right)$$$

Find $$$\int 2 \ln\left(2 x\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \ln{\left(2 x \right)}$$$:

$${\color{red}{\int{2 \ln{\left(2 x \right)} d x}}} = {\color{red}{\left(2 \int{\ln{\left(2 x \right)} d x}\right)}}$$

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

The integral can be rewritten as

$$2 {\color{red}{\int{\ln{\left(2 x \right)} d x}}} = 2 {\color{red}{\int{\frac{\ln{\left(u \right)}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$$2 {\color{red}{\int{\frac{\ln{\left(u \right)}}{2} d u}}} = 2 {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{2}\right)}}$$

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

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

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

The integral becomes

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

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

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

Recall that $$$u=2 x$$$:

$$- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - {\color{red}{\left(2 x\right)}} + {\color{red}{\left(2 x\right)}} \ln{\left({\color{red}{\left(2 x\right)}} \right)}$$

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int 2 \ln\left(2 x\right)\, dx = 2 x \left(\ln\left(x\right) - 1 + \ln\left(2\right)\right) + C$$$A