Integral of $$$\ln\left(\frac{x}{2} - 1\right)$$$

Find $$$\int \ln\left(\frac{x}{2} - 1\right)\, dx$$$.

Let $$$u=\frac{x}{2} - 1$$$.

Then $$$du=\left(\frac{x}{2} - 1\right)^{\prime }dx = \frac{dx}{2}$$$ (steps can be seen »), and we have that $$$dx = 2 du$$$.

So,

$${\color{red}{\int{\ln{\left(\frac{x}{2} - 1 \right)} d x}}} = {\color{red}{\int{2 \ln{\left(u \right)} d u}}}$$

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

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

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 »).

So,

$$2 {\color{red}{\int{\ln{\left(u \right)} d u}}}=2 {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=2 {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$

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

$$2 u \ln{\left(u \right)} - 2 {\color{red}{\int{1 d u}}} = 2 u \ln{\left(u \right)} - 2 {\color{red}{u}}$$

Recall that $$$u=\frac{x}{2} - 1$$$:

$$- 2 {\color{red}{u}} + 2 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - 2 {\color{red}{\left(\frac{x}{2} - 1\right)}} + 2 {\color{red}{\left(\frac{x}{2} - 1\right)}} \ln{\left({\color{red}{\left(\frac{x}{2} - 1\right)}} \right)}$$

Therefore,

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

Simplify:

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

Add the constant of integration (and remove the constant from the expression):

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

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