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

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=6$$$ en $$$f{\left(x \right)} = \frac{1}{1 - x^{2}}$$$:

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

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

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

Integreer termgewijs:

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

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 $$$f{\left(x \right)} = \frac{1}{x + 1}$$$:

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

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 wordt

$$- 6 \int{\frac{1}{2 \left(x - 1\right)} d x} + 3 {\color{red}{\int{\frac{1}{x + 1} d x}}} = - 6 \int{\frac{1}{2 \left(x - 1\right)} d x} + 3 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$- 6 \int{\frac{1}{2 \left(x - 1\right)} d x} + 3 {\color{red}{\int{\frac{1}{u} d u}}} = - 6 \int{\frac{1}{2 \left(x - 1\right)} d x} + 3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - 6 \int{\frac{1}{2 \left(x - 1\right)} d x} = 3 \ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)} - 6 \int{\frac{1}{2 \left(x - 1\right)} d x}$$

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 $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

$$3 \ln{\left(\left|{x + 1}\right| \right)} - 6 {\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}} = 3 \ln{\left(\left|{x + 1}\right| \right)} - 6 {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}$$

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,

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

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

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

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

$$3 \ln{\left(\left|{x + 1}\right| \right)} - 3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 3 \ln{\left(\left|{x + 1}\right| \right)} - 3 \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}$$

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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