Integral of $$$\frac{3}{x - 4}$$$

Find $$$\int \frac{3}{x - 4}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = \frac{1}{x - 4}$$$:

$${\color{red}{\int{\frac{3}{x - 4} d x}}} = {\color{red}{\left(3 \int{\frac{1}{x - 4} d x}\right)}}$$

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

Then $$$du=\left(x - 4\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

$$3 {\color{red}{\int{\frac{1}{x - 4} d x}}} = 3 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$3 {\color{red}{\int{\frac{1}{u} d u}}} = 3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 3 \ln{\left(\left|{{\color{red}{\left(x - 4\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{3}{x - 4} d x} = 3 \ln{\left(\left|{x - 4}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{3}{x - 4} d x} = 3 \ln{\left(\left|{x - 4}\right| \right)}+C$$

$$$\int \frac{3}{x - 4}\, dx = 3 \ln\left(\left|{x - 4}\right|\right) + C$$$A