Integralen av $$$e^{2} \ln\left(x\right)$$$

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

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

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

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 »).

Alltså,

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

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

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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