Integraal van $$$\frac{2 x^{3} - 6 x^{2}}{x - 2}$$$

Bepaal $$$\int \frac{2 x^{3} - 6 x^{2}}{x - 2}\, dx$$$.

Vereenvoudig de integraand:

$${\color{red}{\int{\frac{2 x^{3} - 6 x^{2}}{x - 2} d x}}} = {\color{red}{\int{\frac{2 x^{2} \left(x - 3\right)}{x - 2} d x}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = \frac{x^{2} \left(x - 3\right)}{x - 2}$$$:

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

Aangezien de graad van de teller niet kleiner is dan die van de noemer, voer een staartdeling van polynomen uit (stappen zijn te zien »):

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

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=2$$$:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=4$$$ en $$$f{\left(x \right)} = \frac{1}{x - 2}$$$:

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

Zij $$$u=x - 2$$$.

Dan $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$$\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 {\color{red}{\int{\frac{1}{x - 2} d x}}} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 {\color{red}{\int{\frac{1}{u} d u}}}$$

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=x - 2$$$:

$$\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}$$

Dus,

$$\int{\frac{2 x^{3} - 6 x^{2}}{x - 2} d x} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{x - 2}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{2 x^{3} - 6 x^{2}}{x - 2} d x} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{x - 2}\right| \right)}+C$$

$$$\int \frac{2 x^{3} - 6 x^{2}}{x - 2}\, dx = \left(\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A