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

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

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

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

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

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

The integral becomes

$$\frac{{\color{red}{\int{\frac{1}{x - 2} d x}}}}{2} = \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

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

Therefore,

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

Add the constant of integration:

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

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