Integral of $$$\frac{18}{x \left(x - 3\right)^{2}}$$$

Find $$$\int \frac{18}{x \left(x - 3\right)^{2}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=18$$$ and $$$f{\left(x \right)} = \frac{1}{x \left(x - 3\right)^{2}}$$$:

$${\color{red}{\int{\frac{18}{x \left(x - 3\right)^{2}} d x}}} = {\color{red}{\left(18 \int{\frac{1}{x \left(x - 3\right)^{2}} d x}\right)}}$$

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

$$18 {\color{red}{\int{\frac{1}{x \left(x - 3\right)^{2}} d x}}} = 18 {\color{red}{\int{\left(- \frac{1}{9 \left(x - 3\right)} + \frac{1}{3 \left(x - 3\right)^{2}} + \frac{1}{9 x}\right)d x}}}$$

Integrate term by term:

$$18 {\color{red}{\int{\left(- \frac{1}{9 \left(x - 3\right)} + \frac{1}{3 \left(x - 3\right)^{2}} + \frac{1}{9 x}\right)d x}}} = 18 {\color{red}{\left(\int{\frac{1}{9 x} d x} + \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - \int{\frac{1}{9 \left(x - 3\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{9}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 3}$$$:

$$18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - 18 {\color{red}{\int{\frac{1}{9 \left(x - 3\right)} d x}}} = 18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - 18 {\color{red}{\left(\frac{\int{\frac{1}{x - 3} d x}}{9}\right)}}$$

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 can be rewritten as

$$18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - 2 {\color{red}{\int{\frac{1}{x - 3} d x}}} = 18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - 2 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - 2 {\color{red}{\int{\frac{1}{u} d u}}} = 18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} - 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$- 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x} = - 2 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 18 \int{\frac{1}{3 \left(x - 3\right)^{2}} d x}$$

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

$$- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 18 {\color{red}{\int{\frac{1}{3 \left(x - 3\right)^{2}} d x}}} = - 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 18 {\color{red}{\left(\frac{\int{\frac{1}{\left(x - 3\right)^{2}} d x}}{3}\right)}}$$

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

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

Thus,

$$- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\int{\frac{1}{\left(x - 3\right)^{2}} d x}}} = - 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\int{u^{-2} d u}}}=- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\left(- u^{-1}\right)}}=- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} + 6 {\color{red}{\left(- \frac{1}{u}\right)}}$$

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

$$- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} - 6 {\color{red}{u}}^{-1} = - 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 \int{\frac{1}{9 x} d x} - 6 {\color{red}{\left(x - 3\right)}}^{-1}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{9}$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 {\color{red}{\int{\frac{1}{9 x} d x}}} - \frac{6}{x - 3} = - 2 \ln{\left(\left|{x - 3}\right| \right)} + 18 {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{9}\right)}} - \frac{6}{x - 3}$$

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

$$- 2 \ln{\left(\left|{x - 3}\right| \right)} + 2 {\color{red}{\int{\frac{1}{x} d x}}} - \frac{6}{x - 3} = - 2 \ln{\left(\left|{x - 3}\right| \right)} + 2 {\color{red}{\ln{\left(\left|{x}\right| \right)}}} - \frac{6}{x - 3}$$

Therefore,

$$\int{\frac{18}{x \left(x - 3\right)^{2}} d x} = 2 \ln{\left(\left|{x}\right| \right)} - 2 \ln{\left(\left|{x - 3}\right| \right)} - \frac{6}{x - 3}$$

Simplify:

$$\int{\frac{18}{x \left(x - 3\right)^{2}} d x} = \frac{2 \left(\left(x - 3\right) \left(\ln{\left(\left|{x}\right| \right)} - \ln{\left(\left|{x - 3}\right| \right)}\right) - 3\right)}{x - 3}$$

Add the constant of integration:

$$\int{\frac{18}{x \left(x - 3\right)^{2}} d x} = \frac{2 \left(\left(x - 3\right) \left(\ln{\left(\left|{x}\right| \right)} - \ln{\left(\left|{x - 3}\right| \right)}\right) - 3\right)}{x - 3}+C$$

$$$\int \frac{18}{x \left(x - 3\right)^{2}}\, dx = \frac{2 \left(\left(x - 3\right) \left(\ln\left(\left|{x}\right|\right) - \ln\left(\left|{x - 3}\right|\right)\right) - 3\right)}{x - 3} + C$$$A