Integralen av $$$\ln\left(\frac{x^{n}}{x}\right)$$$ med avseende på $$$x$$$

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

Inmatningen skrivs om: $$$\int{\ln{\left(\frac{x^{n}}{x} \right)} d x}=\int{\left(n - 1\right) \ln{\left(x \right)} d x}$$$.

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

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

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,

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

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

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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