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

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

Låt $$$u=x + 1$$$ vara.

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

Alltså,

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

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

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

Kom ihåg att $$$u=x + 1$$$:

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

Låt $$$u=x - 1$$$ vara.

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

Integralen kan omskrivas som

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

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

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

Kom ihåg att $$$u=x - 1$$$:

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

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

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

Alltså,

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

Lägg till integrationskonstanten:

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

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