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

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=2$$$ och $$$f{\left(x \right)} = \frac{1}{x^{2} \left(1 - x\right)}$$$:

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

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

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

Låt $$$u=x - 1$$$ vara.

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

Integralen blir

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

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

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

Kom ihåg att $$$u=x - 1$$$:

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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