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

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

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

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

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

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

So,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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