Integralen av $$$\frac{1}{\left(x - 2\right) \left(x - 1\right)^{2}}$$$

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

Låt $$$u=x - 2$$$ vara.

Då $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

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

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Kom ihåg att $$$u=x - 2$$$:

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

Låt $$$u=x - 1$$$ vara.

Då $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

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

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Kom ihåg att $$$u=x - 1$$$:

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

Låt $$$u=x - 1$$$ vara.

Då $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Kom ihåg att $$$u=x - 1$$$:

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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