Integral of $$$\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)}$$$

Find $$$\int \frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)}\, dx$$$.

Perform partial fraction decomposition (steps can be seen »):

$${\color{red}{\int{\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)} d x}}} = {\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{5}{3 \left(x - 3\right)} - \frac{2}{3 x}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{5}{3 \left(x - 3\right)} - \frac{2}{3 x}\right)d x}}} = {\color{red}{\left(- \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - \int{\frac{1}{x - 2} d x}\right)}}$$

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

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

Thus,

$$- \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{x - 2} d x}}} = - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$- \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{u} d u}}} = - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} = - \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)} - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{2}{3}$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{2}{3 x} d x}}} = - \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\left(\frac{2 \int{\frac{1}{x} d x}}{3}\right)}}$$

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

$$- \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - \frac{2 {\color{red}{\int{\frac{1}{x} d x}}}}{3} = - \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - \frac{2 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{3}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{5}{3}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 3}$$$:

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + {\color{red}{\int{\frac{5}{3 \left(x - 3\right)} d x}}} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + {\color{red}{\left(\frac{5 \int{\frac{1}{x - 3} d x}}{3}\right)}}$$

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

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

The integral becomes

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\int{\frac{1}{x - 3} d x}}}}{3} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{3}$$

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

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{3} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{3}$$

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

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{3} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)}}{3}$$

Therefore,

$$\int{\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)} d x} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} + \frac{5 \ln{\left(\left|{x - 3}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)} d x} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} + \frac{5 \ln{\left(\left|{x - 3}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)}+C$$

$$$\int \frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)}\, dx = \left(- \frac{2 \ln\left(\left|{x}\right|\right)}{3} + \frac{5 \ln\left(\left|{x - 3}\right|\right)}{3} - \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A