Integralen av $$$\frac{x^{4}}{x^{2} - 1}$$$

Bestäm $$$\int \frac{x^{4}}{x^{2} - 1}\, dx$$$.

Eftersom graden hos täljaren inte är mindre än graden hos nämnaren, utför polynomdivision (stegen kan ses »):

$${\color{red}{\int{\frac{x^{4}}{x^{2} - 1} d x}}} = {\color{red}{\int{\left(x^{2} + 1 + \frac{1}{x^{2} - 1}\right)d x}}}$$

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=1$$$:

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

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

$$x + \int{\frac{1}{x^{2} - 1} d x} + {\color{red}{\int{x^{2} d x}}}=x + \int{\frac{1}{x^{2} - 1} d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=x + \int{\frac{1}{x^{2} - 1} d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Utför partialbråksuppdelning (stegen kan ses »):

$$\frac{x^{3}}{3} + x + {\color{red}{\int{\frac{1}{x^{2} - 1} d x}}} = \frac{x^{3}}{3} + x + {\color{red}{\int{\left(- \frac{1}{2 \left(x + 1\right)} + \frac{1}{2 \left(x - 1\right)}\right)d x}}}$$

Integrera termvis:

$$\frac{x^{3}}{3} + x + {\color{red}{\int{\left(- \frac{1}{2 \left(x + 1\right)} + \frac{1}{2 \left(x - 1\right)}\right)d x}}} = \frac{x^{3}}{3} + x + {\color{red}{\left(\int{\frac{1}{2 \left(x - 1\right)} d x} - \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}$$

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

$$\frac{x^{3}}{3} + x - \int{\frac{1}{2 \left(x + 1\right)} d x} + {\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}} = \frac{x^{3}}{3} + x - \int{\frac{1}{2 \left(x + 1\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}$$

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,

$$\frac{x^{3}}{3} + x - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{2} = \frac{x^{3}}{3} + x - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$\frac{x^{3}}{3} + x - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{x^{3}}{3} + x - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Kom ihåg att $$$u=x - 1$$$:

$$\frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(x + 1\right)} d x} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(x + 1\right)} d x}$$

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

$$\frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - {\color{red}{\int{\frac{1}{2 \left(x + 1\right)} d x}}} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{2}\right)}}$$

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

$$\frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{x + 1} d x}}}}{2} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$\frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Kom ihåg att $$$u=x + 1$$$:

$$\frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2}$$

Alltså,

$$\int{\frac{x^{4}}{x^{2} - 1} d x} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}$$

Lägg till integrationskonstanten:

$$\int{\frac{x^{4}}{x^{2} - 1} d x} = \frac{x^{3}}{3} + x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}+C$$

$$$\int \frac{x^{4}}{x^{2} - 1}\, dx = \left(\frac{x^{3}}{3} + x + \frac{\ln\left(\left|{x - 1}\right|\right)}{2} - \frac{\ln\left(\left|{x + 1}\right|\right)}{2}\right) + C$$$A