Integral of $$$\frac{x + 6}{x - 6}$$$

Find $$$\int \frac{x + 6}{x - 6}\, dx$$$.

Let $$$u=x - 6$$$.

Then $$$du=\left(x - 6\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$${\color{red}{\int{\frac{x + 6}{x - 6} d x}}} = {\color{red}{\int{\frac{u + 12}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{u + 12}{u} d u}}} = {\color{red}{\int{\left(1 + \frac{12}{u}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(1 + \frac{12}{u}\right)d u}}} = {\color{red}{\left(\int{1 d u} + \int{\frac{12}{u} d u}\right)}}$$

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

$$\int{\frac{12}{u} d u} + {\color{red}{\int{1 d u}}} = \int{\frac{12}{u} d u} + {\color{red}{u}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=12$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$u + {\color{red}{\int{\frac{12}{u} d u}}} = u + {\color{red}{\left(12 \int{\frac{1}{u} d u}\right)}}$$

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

$$u + 12 {\color{red}{\int{\frac{1}{u} d u}}} = u + 12 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=x - 6$$$:

$$12 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + {\color{red}{u}} = 12 \ln{\left(\left|{{\color{red}{\left(x - 6\right)}}}\right| \right)} + {\color{red}{\left(x - 6\right)}}$$

Therefore,

$$\int{\frac{x + 6}{x - 6} d x} = x + 12 \ln{\left(\left|{x - 6}\right| \right)} - 6$$

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

$$\int{\frac{x + 6}{x - 6} d x} = x + 12 \ln{\left(\left|{x - 6}\right| \right)}+C$$

$$$\int \frac{x + 6}{x - 6}\, dx = \left(x + 12 \ln\left(\left|{x - 6}\right|\right)\right) + C$$$A