Integraal van $$$\frac{x^{6} - 1}{x^{2} + 1}$$$

Bepaal $$$\int \frac{x^{6} - 1}{x^{2} + 1}\, dx$$$.

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

$${\color{red}{\int{\frac{x^{6} - 1}{x^{2} + 1} d x}}} = {\color{red}{\int{\left(x^{4} - x^{2} + 1 - \frac{2}{x^{2} + 1}\right)d x}}}$$

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=4$$$:

$$x - \int{x^{2} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\int{x^{4} d x}}}=x - \int{x^{2} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=x - \int{x^{2} d x} - \int{\frac{2}{x^{2} + 1} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

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

$$\frac{x^{5}}{5} + x - \int{\frac{2}{x^{2} + 1} d x} - {\color{red}{\int{x^{2} d x}}}=\frac{x^{5}}{5} + x - \int{\frac{2}{x^{2} + 1} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\frac{x^{5}}{5} + x - \int{\frac{2}{x^{2} + 1} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

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

$$\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - {\color{red}{\int{\frac{2}{x^{2} + 1} d x}}} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - {\color{red}{\left(2 \int{\frac{1}{x^{2} + 1} d x}\right)}}$$

De integraal van $$$\frac{1}{x^{2} + 1}$$$ is $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

$$\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 {\color{red}{\int{\frac{1}{x^{2} + 1} d x}}} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 {\color{red}{\operatorname{atan}{\left(x \right)}}}$$

Dus,

$$\int{\frac{x^{6} - 1}{x^{2} + 1} d x} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 \operatorname{atan}{\left(x \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{x^{6} - 1}{x^{2} + 1} d x} = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 \operatorname{atan}{\left(x \right)}+C$$

$$$\int \frac{x^{6} - 1}{x^{2} + 1}\, dx = \left(\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - 2 \operatorname{atan}{\left(x \right)}\right) + C$$$A