Integralen av $$$\frac{\ln\left(x\right)}{5}$$$

Bestäm $$$\int \frac{\ln\left(x\right)}{5}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{5}$$$ och $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\ln{\left(x \right)}}{5} d x}}} = {\color{red}{\left(\frac{\int{\ln{\left(x \right)} d x}}{5}\right)}}$$

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

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

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

Integralen kan omskrivas som

$$\frac{{\color{red}{\int{\ln{\left(x \right)} d x}}}}{5}=\frac{{\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}}{5}=\frac{{\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}}{5}$$

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

$$\frac{x \ln{\left(x \right)}}{5} - \frac{{\color{red}{\int{1 d x}}}}{5} = \frac{x \ln{\left(x \right)}}{5} - \frac{{\color{red}{x}}}{5}$$

Alltså,

$$\int{\frac{\ln{\left(x \right)}}{5} d x} = \frac{x \ln{\left(x \right)}}{5} - \frac{x}{5}$$

Förenkla:

$$\int{\frac{\ln{\left(x \right)}}{5} d x} = \frac{x \left(\ln{\left(x \right)} - 1\right)}{5}$$

Lägg till integrationskonstanten:

$$\int{\frac{\ln{\left(x \right)}}{5} d x} = \frac{x \left(\ln{\left(x \right)} - 1\right)}{5}+C$$

$$$\int \frac{\ln\left(x\right)}{5}\, dx = \frac{x \left(\ln\left(x\right) - 1\right)}{5} + C$$$A