Integral of $$$\frac{8 x^{2} - 64 x}{x^{3} - 4 x}$$$

Find $$$\int \frac{8 x^{2} - 64 x}{x^{3} - 4 x}\, dx$$$.

Simplify the integrand:

$${\color{red}{\int{\frac{8 x^{2} - 64 x}{x^{3} - 4 x} d x}}} = {\color{red}{\int{\frac{8 x - 64}{x^{2} - 4} d x}}}$$

Simplify the integrand:

$${\color{red}{\int{\frac{8 x - 64}{x^{2} - 4} d x}}} = {\color{red}{\int{\frac{8 \left(x - 8\right)}{x^{2} - 4} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=8$$$ and $$$f{\left(x \right)} = \frac{x - 8}{x^{2} - 4}$$$:

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

Split the fraction:

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

Integrate term by term:

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

Let $$$u=x^{2} - 4$$$.

Then $$$du=\left(x^{2} - 4\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.

The integral becomes

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

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

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

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

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

Recall that $$$u=x^{2} - 4$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-8$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2} - 4}$$$:

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

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

$$4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 {\color{red}{\int{\frac{1}{x^{2} - 4} d x}}} = 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 {\color{red}{\int{\left(- \frac{1}{4 \left(x + 2\right)} + \frac{1}{4 \left(x - 2\right)}\right)d x}}}$$

Integrate term by term:

$$4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 {\color{red}{\int{\left(- \frac{1}{4 \left(x + 2\right)} + \frac{1}{4 \left(x - 2\right)}\right)d x}}} = 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 {\color{red}{\left(\int{\frac{1}{4 \left(x - 2\right)} d x} - \int{\frac{1}{4 \left(x + 2\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{x + 2}$$$:

$$4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} + 64 {\color{red}{\int{\frac{1}{4 \left(x + 2\right)} d x}}} = 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} + 64 {\color{red}{\left(\frac{\int{\frac{1}{x + 2} d x}}{4}\right)}}$$

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,

$$4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} + 16 {\color{red}{\int{\frac{1}{x + 2} d x}}} = 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} + 16 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} + 16 {\color{red}{\int{\frac{1}{u} d u}}} = 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} + 16 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$4 \ln{\left(\left|{x^{2} - 4}\right| \right)} + 16 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x} = 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} + 16 \ln{\left(\left|{{\color{red}{\left(x + 2\right)}}}\right| \right)} - 64 \int{\frac{1}{4 \left(x - 2\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 2}$$$:

$$16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 {\color{red}{\int{\frac{1}{4 \left(x - 2\right)} d x}}} = 16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 64 {\color{red}{\left(\frac{\int{\frac{1}{x - 2} d x}}{4}\right)}}$$

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

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

The integral can be rewritten as

$$16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 16 {\color{red}{\int{\frac{1}{x - 2} d x}}} = 16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 16 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 16 {\color{red}{\int{\frac{1}{u} d u}}} = 16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 16 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 16 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)} - 16 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{8 x^{2} - 64 x}{x^{3} - 4 x} d x} = - 16 \ln{\left(\left|{x - 2}\right| \right)} + 16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{8 x^{2} - 64 x}{x^{3} - 4 x} d x} = - 16 \ln{\left(\left|{x - 2}\right| \right)} + 16 \ln{\left(\left|{x + 2}\right| \right)} + 4 \ln{\left(\left|{x^{2} - 4}\right| \right)}+C$$

$$$\int \frac{8 x^{2} - 64 x}{x^{3} - 4 x}\, dx = \left(- 16 \ln\left(\left|{x - 2}\right|\right) + 16 \ln\left(\left|{x + 2}\right|\right) + 4 \ln\left(\left|{x^{2} - 4}\right|\right)\right) + C$$$A