Integraal van $$$\frac{10}{100 - x^{2}}$$$

Bepaal $$$\int \frac{10}{100 - x^{2}}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=10$$$ en $$$f{\left(x \right)} = \frac{1}{100 - x^{2}}$$$:

$${\color{red}{\int{\frac{10}{100 - x^{2}} d x}}} = {\color{red}{\left(10 \int{\frac{1}{100 - x^{2}} d x}\right)}}$$

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

$$10 {\color{red}{\int{\frac{1}{100 - x^{2}} d x}}} = 10 {\color{red}{\int{\left(\frac{1}{20 \left(x + 10\right)} - \frac{1}{20 \left(x - 10\right)}\right)d x}}}$$

Integreer termgewijs:

$$10 {\color{red}{\int{\left(\frac{1}{20 \left(x + 10\right)} - \frac{1}{20 \left(x - 10\right)}\right)d x}}} = 10 {\color{red}{\left(- \int{\frac{1}{20 \left(x - 10\right)} d x} + \int{\frac{1}{20 \left(x + 10\right)} d x}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{20}$$$ en $$$f{\left(x \right)} = \frac{1}{x - 10}$$$:

$$10 \int{\frac{1}{20 \left(x + 10\right)} d x} - 10 {\color{red}{\int{\frac{1}{20 \left(x - 10\right)} d x}}} = 10 \int{\frac{1}{20 \left(x + 10\right)} d x} - 10 {\color{red}{\left(\frac{\int{\frac{1}{x - 10} d x}}{20}\right)}}$$

Zij $$$u=x - 10$$$.

Dan $$$du=\left(x - 10\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

$$10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x - 10} d x}}}}{2} = 10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = 10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

We herinneren eraan dat $$$u=x - 10$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + 10 \int{\frac{1}{20 \left(x + 10\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x - 10\right)}}}\right| \right)}}{2} + 10 \int{\frac{1}{20 \left(x + 10\right)} d x}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{20}$$$ en $$$f{\left(x \right)} = \frac{1}{x + 10}$$$:

$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + 10 {\color{red}{\int{\frac{1}{20 \left(x + 10\right)} d x}}} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + 10 {\color{red}{\left(\frac{\int{\frac{1}{x + 10} d x}}{20}\right)}}$$

Zij $$$u=x + 10$$$.

Dan $$$du=\left(x + 10\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

De integraal kan worden herschreven als

$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{x + 10} d x}}}}{2} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

We herinneren eraan dat $$$u=x + 10$$$:

$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(x + 10\right)}}}\right| \right)}}{2}$$

Dus,

$$\int{\frac{10}{100 - x^{2}} d x} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{\ln{\left(\left|{x + 10}\right| \right)}}{2}$$

Vereenvoudig:

$$\int{\frac{10}{100 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 10}\right| \right)} + \ln{\left(\left|{x + 10}\right| \right)}}{2}$$

Voeg de integratieconstante toe:

$$\int{\frac{10}{100 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 10}\right| \right)} + \ln{\left(\left|{x + 10}\right| \right)}}{2}+C$$

$$$\int \frac{10}{100 - x^{2}}\, dx = \frac{- \ln\left(\left|{x - 10}\right|\right) + \ln\left(\left|{x + 10}\right|\right)}{2} + C$$$A