Integraal van $$$\frac{1}{4 - 9 x^{2}}$$$

Bepaal $$$\int \frac{1}{4 - 9 x^{2}}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{4 - 9 x^{2}} d x}}} = {\color{red}{\int{\left(\frac{1}{4 \left(3 x + 2\right)} - \frac{1}{4 \left(3 x - 2\right)}\right)d x}}}$$

Integreer termgewijs:

$${\color{red}{\int{\left(\frac{1}{4 \left(3 x + 2\right)} - \frac{1}{4 \left(3 x - 2\right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{4 \left(3 x - 2\right)} d x} + \int{\frac{1}{4 \left(3 x + 2\right)} d x}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$:

$$\int{\frac{1}{4 \left(3 x + 2\right)} d x} - {\color{red}{\int{\frac{1}{4 \left(3 x - 2\right)} d x}}} = \int{\frac{1}{4 \left(3 x + 2\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{3 x - 2} d x}}{4}\right)}}$$

Zij $$$u=3 x - 2$$$.

Dan $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{3}$$$.

Dus,

$$\int{\frac{1}{4 \left(3 x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 x - 2} d x}}}}{4} = \int{\frac{1}{4 \left(3 x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$\int{\frac{1}{4 \left(3 x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4} = \int{\frac{1}{4 \left(3 x + 2\right)} d x} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{4}$$

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

$$\int{\frac{1}{4 \left(3 x + 2\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{12} = \int{\frac{1}{4 \left(3 x + 2\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{12}$$

We herinneren eraan dat $$$u=3 x - 2$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{12} + \int{\frac{1}{4 \left(3 x + 2\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}}{12} + \int{\frac{1}{4 \left(3 x + 2\right)} d x}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = \frac{1}{3 x + 2}$$$:

$$- \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + {\color{red}{\int{\frac{1}{4 \left(3 x + 2\right)} d x}}} = - \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + {\color{red}{\left(\frac{\int{\frac{1}{3 x + 2} d x}}{4}\right)}}$$

Zij $$$u=3 x + 2$$$.

Dan $$$du=\left(3 x + 2\right)^{\prime }dx = 3 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{3}$$$.

De integraal kan worden herschreven als

$$- \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 x + 2} d x}}}}{4} = - \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$- \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{3 u} d u}}}}{4} = - \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{4}$$

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

$$- \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{12} = - \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{12}$$

We herinneren eraan dat $$$u=3 x + 2$$$:

$$- \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{12} = - \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{{\color{red}{\left(3 x + 2\right)}}}\right| \right)}}{12}$$

Dus,

$$\int{\frac{1}{4 - 9 x^{2}} d x} = - \frac{\ln{\left(\left|{3 x - 2}\right| \right)}}{12} + \frac{\ln{\left(\left|{3 x + 2}\right| \right)}}{12}$$

Vereenvoudig:

$$\int{\frac{1}{4 - 9 x^{2}} d x} = \frac{- \ln{\left(\left|{3 x - 2}\right| \right)} + \ln{\left(\left|{3 x + 2}\right| \right)}}{12}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{4 - 9 x^{2}} d x} = \frac{- \ln{\left(\left|{3 x - 2}\right| \right)} + \ln{\left(\left|{3 x + 2}\right| \right)}}{12}+C$$

$$$\int \frac{1}{4 - 9 x^{2}}\, dx = \frac{- \ln\left(\left|{3 x - 2}\right|\right) + \ln\left(\left|{3 x + 2}\right|\right)}{12} + C$$$A