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

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

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

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

Integreer termgewijs:

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

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

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

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

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

Voer een ontbinding in partiële breuken uit (stappen zijn te zien »):

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

Integreer termgewijs:

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

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

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

Zij $$$u=x - 1$$$.

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

De integraal kan worden herschreven als

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

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

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

We herinneren eraan dat $$$u=x - 1$$$:

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

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

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

Zij $$$u=x + 1$$$.

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

Dus,

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

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

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

We herinneren eraan dat $$$u=x + 1$$$:

$$x + \ln{\left(\left|{x - 1}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x + \ln{\left(\left|{x - 1}\right| \right)} - \ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}$$

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \frac{x^{2} + 1}{x^{2} - 1}\, dx = \left(x + \ln\left(\left|{x - 1}\right|\right) - \ln\left(\left|{x + 1}\right|\right)\right) + C$$$A