Integralen av $$$\frac{- 2 \ln\left(x\right) - 4}{x}$$$

Bestäm $$$\int \frac{- 2 \ln\left(x\right) - 4}{x}\, dx$$$.

Förenkla integranden:

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

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

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

Expand the expression:

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

Integrera termvis:

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

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

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

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

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

Låt $$$u=\ln{\left(x \right)}$$$ vara.

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

Alltså,

$$- 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\int{\frac{\ln{\left(x \right)}}{x} d x}}} = - 4 \ln{\left(\left|{x}\right| \right)} - 2 {\color{red}{\int{u d u}}}$$

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

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

Kom ihåg att $$$u=\ln{\left(x \right)}$$$:

$$- 4 \ln{\left(\left|{x}\right| \right)} - {\color{red}{u}}^{2} = - 4 \ln{\left(\left|{x}\right| \right)} - {\color{red}{\ln{\left(x \right)}}}^{2}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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