Integral of $$$\frac{1}{x - 3}$$$

Find $$$\int \frac{1}{x - 3}\, dx$$$.

Let $$$u=x - 3$$$.

Then $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

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

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

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

Recall that $$$u=x - 3$$$:

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

Therefore,

$$\int{\frac{1}{x - 3} d x} = \ln{\left(\left|{x - 3}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{1}{x - 3} d x} = \ln{\left(\left|{x - 3}\right| \right)}+C$$

$$$\int \frac{1}{x - 3}\, dx = \ln\left(\left|{x - 3}\right|\right) + C$$$A