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

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

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

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

För integralen $$$\int{\ln{\left(1 - x^{2} \right)} d x}$$$, använd partiell integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Låt $$$\operatorname{u}=\ln{\left(1 - x^{2} \right)}$$$ och $$$\operatorname{dv}=dx$$$.

Då gäller $$$\operatorname{du}=\left(\ln{\left(1 - x^{2} \right)}\right)^{\prime }dx=\frac{2 x}{x^{2} - 1} dx$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d x}=x$$$ (stegen kan ses »).

Integralen kan omskrivas som

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

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

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

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

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

Integrera termvis:

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

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

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

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,

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

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

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

Kom ihåg att $$$u=x - 1$$$:

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

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

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

Låt $$$u=x + 1$$$ vara.

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

Alltså,

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

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

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

Kom ihåg att $$$u=x + 1$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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