Integralen av $$$- 21 x - 3 \ln\left(- x\right)$$$

Bestäm $$$\int \left(- 21 x - 3 \ln\left(- x\right)\right)\, dx$$$.

Integrera termvis:

$${\color{red}{\int{\left(- 21 x - 3 \ln{\left(- x \right)}\right)d x}}} = {\color{red}{\left(- \int{21 x d x} - \int{3 \ln{\left(- x \right)} d x}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=21$$$ och $$$f{\left(x \right)} = x$$$:

$$- \int{3 \ln{\left(- x \right)} d x} - {\color{red}{\int{21 x d x}}} = - \int{3 \ln{\left(- x \right)} d x} - {\color{red}{\left(21 \int{x d x}\right)}}$$

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

$$- \int{3 \ln{\left(- x \right)} d x} - 21 {\color{red}{\int{x d x}}}=- \int{3 \ln{\left(- x \right)} d x} - 21 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- \int{3 \ln{\left(- x \right)} d x} - 21 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=3$$$ och $$$f{\left(x \right)} = \ln{\left(- x \right)}$$$:

$$- \frac{21 x^{2}}{2} - {\color{red}{\int{3 \ln{\left(- x \right)} d x}}} = - \frac{21 x^{2}}{2} - {\color{red}{\left(3 \int{\ln{\left(- x \right)} d x}\right)}}$$

Låt $$$u=- x$$$ vara.

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

Alltså,

$$- \frac{21 x^{2}}{2} - 3 {\color{red}{\int{\ln{\left(- x \right)} d x}}} = - \frac{21 x^{2}}{2} - 3 {\color{red}{\int{\left(- \ln{\left(u \right)}\right)d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=-1$$$ och $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$$- \frac{21 x^{2}}{2} - 3 {\color{red}{\int{\left(- \ln{\left(u \right)}\right)d u}}} = - \frac{21 x^{2}}{2} - 3 {\color{red}{\left(- \int{\ln{\left(u \right)} d u}\right)}}$$

För integralen $$$\int{\ln{\left(u \right)} d u}$$$, använd partiell integration $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Låt $$$\operatorname{g}=\ln{\left(u \right)}$$$ och $$$\operatorname{dv}=du$$$.

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

Alltså,

$$- \frac{21 x^{2}}{2} + 3 {\color{red}{\int{\ln{\left(u \right)} d u}}}=- \frac{21 x^{2}}{2} + 3 {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=- \frac{21 x^{2}}{2} + 3 {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

$$3 u \ln{\left(u \right)} - \frac{21 x^{2}}{2} - 3 {\color{red}{\int{1 d u}}} = 3 u \ln{\left(u \right)} - \frac{21 x^{2}}{2} - 3 {\color{red}{u}}$$

Kom ihåg att $$$u=- x$$$:

$$- \frac{21 x^{2}}{2} - 3 {\color{red}{u}} + 3 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = - \frac{21 x^{2}}{2} - 3 {\color{red}{\left(- x\right)}} + 3 {\color{red}{\left(- x\right)}} \ln{\left({\color{red}{\left(- x\right)}} \right)}$$

Alltså,

$$\int{\left(- 21 x - 3 \ln{\left(- x \right)}\right)d x} = - \frac{21 x^{2}}{2} - 3 x \ln{\left(- x \right)} + 3 x$$

Förenkla:

$$\int{\left(- 21 x - 3 \ln{\left(- x \right)}\right)d x} = \frac{3 x \left(- 7 x - 2 \ln{\left(- x \right)} + 2\right)}{2}$$

Lägg till integrationskonstanten:

$$\int{\left(- 21 x - 3 \ln{\left(- x \right)}\right)d x} = \frac{3 x \left(- 7 x - 2 \ln{\left(- x \right)} + 2\right)}{2}+C$$

$$$\int \left(- 21 x - 3 \ln\left(- x\right)\right)\, dx = \frac{3 x \left(- 7 x - 2 \ln\left(- x\right) + 2\right)}{2} + C$$$A