Integraal van $$$\frac{1}{2 \left(1 - x^{2}\right)}$$$

Bepaal $$$\int \frac{1}{2 \left(1 - x^{2}\right)}\, dx$$$.

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 $$$f{\left(x \right)} = \frac{1}{1 - x^{2}}$$$:

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

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

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

Integreer termgewijs:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{4}$$

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

$$\int{\frac{1}{2 \left(1 - x^{2}\right)} d x} = \frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{4}+C$$

$$$\int \frac{1}{2 \left(1 - x^{2}\right)}\, dx = \frac{- \ln\left(\left|{x - 1}\right|\right) + \ln\left(\left|{x + 1}\right|\right)}{4} + C$$$A