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

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

Låt $$$u=x - 1$$$ vara.

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

Alltså,

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

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

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

Kom ihåg att $$$u=x - 1$$$:

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

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

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

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

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

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

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

Kom ihåg att $$$u=x + 1$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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