Integraal van $$$\frac{2 x^{3}}{x^{2} - 9}$$$

Bepaal $$$\int \frac{2 x^{3}}{x^{2} - 9}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = \frac{x^{3}}{x^{2} - 9}$$$:

$${\color{red}{\int{\frac{2 x^{3}}{x^{2} - 9} d x}}} = {\color{red}{\left(2 \int{\frac{x^{3}}{x^{2} - 9} d x}\right)}}$$

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

$$2 {\color{red}{\int{\frac{x^{3}}{x^{2} - 9} d x}}} = 2 {\color{red}{\int{\left(x + \frac{9 x}{x^{2} - 9}\right)d x}}}$$

Integreer termgewijs:

$$2 {\color{red}{\int{\left(x + \frac{9 x}{x^{2} - 9}\right)d x}}} = 2 {\color{red}{\left(\int{x d x} + \int{\frac{9 x}{x^{2} - 9} d x}\right)}}$$

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

$$2 \int{\frac{9 x}{x^{2} - 9} d x} + 2 {\color{red}{\int{x d x}}}=2 \int{\frac{9 x}{x^{2} - 9} d x} + 2 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=2 \int{\frac{9 x}{x^{2} - 9} d x} + 2 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Zij $$$u=x^{2} - 9$$$.

Dan $$$du=\left(x^{2} - 9\right)^{\prime }dx = 2 x dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$x dx = \frac{du}{2}$$$.

Dus,

$$x^{2} + 2 {\color{red}{\int{\frac{9 x}{x^{2} - 9} d x}}} = x^{2} + 2 {\color{red}{\int{\frac{9}{2 u} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{9}{2}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$x^{2} + 2 {\color{red}{\int{\frac{9}{2 u} d u}}} = x^{2} + 2 {\color{red}{\left(\frac{9 \int{\frac{1}{u} d u}}{2}\right)}}$$

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

$$x^{2} + 9 {\color{red}{\int{\frac{1}{u} d u}}} = x^{2} + 9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

We herinneren eraan dat $$$u=x^{2} - 9$$$:

$$x^{2} + 9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x^{2} + 9 \ln{\left(\left|{{\color{red}{\left(x^{2} - 9\right)}}}\right| \right)}$$

Dus,

$$\int{\frac{2 x^{3}}{x^{2} - 9} d x} = x^{2} + 9 \ln{\left(\left|{x^{2} - 9}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{\frac{2 x^{3}}{x^{2} - 9} d x} = x^{2} + 9 \ln{\left(\left|{x^{2} - 9}\right| \right)}+C$$

$$$\int \frac{2 x^{3}}{x^{2} - 9}\, dx = \left(x^{2} + 9 \ln\left(\left|{x^{2} - 9}\right|\right)\right) + C$$$A