Integraal van $$$\frac{1}{2 \left(4 - x^{2}\right)}$$$

Bepaal $$$\int \frac{1}{2 \left(4 - x^{2}\right)}\, dx$$$.

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 $$$f{\left(x \right)} = \frac{1}{4 - x^{2}}$$$:

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

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

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

Integreer termgewijs:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = \frac{1}{x - 2}$$$:

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

Zij $$$u=x - 2$$$.

Dan $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$$\frac{\int{\frac{1}{4 \left(x + 2\right)} d x}}{2} - \frac{{\color{red}{\int{\frac{1}{x - 2} d x}}}}{8} = \frac{\int{\frac{1}{4 \left(x + 2\right)} d x}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$\frac{\int{\frac{1}{4 \left(x + 2\right)} d x}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{\int{\frac{1}{4 \left(x + 2\right)} d x}}{2} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

We herinneren eraan dat $$$u=x - 2$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{\int{\frac{1}{4 \left(x + 2\right)} d x}}{2} = - \frac{\ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}}{8} + \frac{\int{\frac{1}{4 \left(x + 2\right)} d x}}{2}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = \frac{1}{x + 2}$$$:

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{4 \left(x + 2\right)} d x}}}}{2} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{x + 2} d x}}{4}\right)}}}{2}$$

Zij $$$u=x + 2$$$.

Dan $$$du=\left(x + 2\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{x + 2} d x}}}}{8} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

We herinneren eraan dat $$$u=x + 2$$$:

$$- \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{\ln{\left(\left|{{\color{red}{\left(x + 2\right)}}}\right| \right)}}{8}$$

Dus,

$$\int{\frac{1}{2 \left(4 - x^{2}\right)} d x} = - \frac{\ln{\left(\left|{x - 2}\right| \right)}}{8} + \frac{\ln{\left(\left|{x + 2}\right| \right)}}{8}$$

Vereenvoudig:

$$\int{\frac{1}{2 \left(4 - x^{2}\right)} d x} = \frac{- \ln{\left(\left|{x - 2}\right| \right)} + \ln{\left(\left|{x + 2}\right| \right)}}{8}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{2 \left(4 - x^{2}\right)} d x} = \frac{- \ln{\left(\left|{x - 2}\right| \right)} + \ln{\left(\left|{x + 2}\right| \right)}}{8}+C$$

$$$\int \frac{1}{2 \left(4 - x^{2}\right)}\, dx = \frac{- \ln\left(\left|{x - 2}\right|\right) + \ln\left(\left|{x + 2}\right|\right)}{8} + C$$$A