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

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

Simplify the integrand:

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

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

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

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

$$2 {\color{red}{\int{\frac{x^{3} - 1}{x - 2} d x}}} = 2 {\color{red}{\int{\left(x^{2} + 2 x + 4 + \frac{7}{x - 2}\right)d x}}}$$

Integrate term by term:

$$2 {\color{red}{\int{\left(x^{2} + 2 x + 4 + \frac{7}{x - 2}\right)d x}}} = 2 {\color{red}{\left(\int{4 d x} + \int{2 x d x} + \int{x^{2} d x} + \int{\frac{7}{x - 2} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=4$$$:

$$2 \int{2 x d x} + 2 \int{x^{2} d x} + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\int{4 d x}}} = 2 \int{2 x d x} + 2 \int{x^{2} d x} + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\left(4 x\right)}}$$

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

$$8 x + 2 \int{2 x d x} + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\int{x^{2} d x}}}=8 x + 2 \int{2 x d x} + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=8 x + 2 \int{2 x d x} + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

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

$$\frac{2 x^{3}}{3} + 8 x + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\int{2 x d x}}} = \frac{2 x^{3}}{3} + 8 x + 2 \int{\frac{7}{x - 2} d x} + 2 {\color{red}{\left(2 \int{x d x}\right)}}$$

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

$$\frac{2 x^{3}}{3} + 8 x + 2 \int{\frac{7}{x - 2} d x} + 4 {\color{red}{\int{x d x}}}=\frac{2 x^{3}}{3} + 8 x + 2 \int{\frac{7}{x - 2} d x} + 4 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{2 x^{3}}{3} + 8 x + 2 \int{\frac{7}{x - 2} d x} + 4 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=7$$$ and $$$f{\left(x \right)} = \frac{1}{x - 2}$$$:

$$\frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 2 {\color{red}{\int{\frac{7}{x - 2} d x}}} = \frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 2 {\color{red}{\left(7 \int{\frac{1}{x - 2} d x}\right)}}$$

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

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

The integral can be rewritten as

$$\frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 {\color{red}{\int{\frac{1}{x - 2} d x}}} = \frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$\frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 {\color{red}{\int{\frac{1}{u} d u}}} = \frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$\frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}$$

Therefore,

$$\int{\frac{2 x^{3} - 2}{x - 2} d x} = \frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 \ln{\left(\left|{x - 2}\right| \right)}$$

Add the constant of integration:

$$\int{\frac{2 x^{3} - 2}{x - 2} d x} = \frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 \ln{\left(\left|{x - 2}\right| \right)}+C$$

$$$\int \frac{2 x^{3} - 2}{x - 2}\, dx = \left(\frac{2 x^{3}}{3} + 2 x^{2} + 8 x + 14 \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A