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

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

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

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

Integreer termgewijs:

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

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

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

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

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

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,

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

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

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

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

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

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

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

De integraal kan worden herschreven als

$$\int{\frac{2}{x - 1} d x} - 2 {\color{red}{\int{\frac{1}{x - 2} d x}}} - \frac{1}{x - 1} - \frac{1}{x - 2} = \int{\frac{2}{x - 1} d x} - 2 {\color{red}{\int{\frac{1}{u} d u}}} - \frac{1}{x - 1} - \frac{1}{x - 2}$$

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

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

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

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

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

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

Zij $$$u=x - 1$$$.

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

Dus,

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

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

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

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

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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