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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{x}{\left(x - 2\right)^{2}}$$$:

$${\color{red}{\int{\frac{4 x}{\left(x - 2\right)^{2}} d x}}} = {\color{red}{\left(4 \int{\frac{x}{\left(x - 2\right)^{2}} d x}\right)}}$$

Rewrite the numerator of the integrand as $$$x=x - 2+2$$$ and split the fraction:

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

Integrate term by term:

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

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,

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

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

$$4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} + 4 {\color{red}{\int{\frac{1}{u} d u}}} = 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} + 4 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$4 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x} = 4 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)} + 4 \int{\frac{2}{\left(x - 2\right)^{2}} d x}$$

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

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

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,

$$4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{\frac{1}{\left(x - 2\right)^{2}} d x}}} = 4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

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

$$4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\int{u^{-2} d u}}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\left(- u^{-1}\right)}}=4 \ln{\left(\left|{x - 2}\right| \right)} + 8 {\color{red}{\left(- \frac{1}{u}\right)}}$$

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

$$4 \ln{\left(\left|{x - 2}\right| \right)} - 8 {\color{red}{u}}^{-1} = 4 \ln{\left(\left|{x - 2}\right| \right)} - 8 {\color{red}{\left(x - 2\right)}}^{-1}$$

Therefore,

$$\int{\frac{4 x}{\left(x - 2\right)^{2}} d x} = 4 \ln{\left(\left|{x - 2}\right| \right)} - \frac{8}{x - 2}$$

Simplify:

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

Add the constant of integration:

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

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