Integralen av $$$\frac{2}{1 - x^{2}}$$$

Bestäm $$$\int \frac{2}{1 - x^{2}}\, dx$$$.

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

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

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

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

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

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

Kom ihåg att $$$u=x + 1$$$:

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

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

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

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

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

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

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

Kom ihåg att $$$u=x - 1$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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