Integralen av $$$\ln\left(x^{4}\right) - \ln\left(y^{3}\right)$$$ med avseende på $$$x$$$

Bestäm $$$\int \left(\ln\left(x^{4}\right) - 3 \ln\left(y\right)\right)\, dx$$$.

Inmatningen skrivs om: $$$\int{\left(\ln{\left(x^{4} \right)} - \ln{\left(y^{3} \right)}\right)d x}=\int{\left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)}\right)d x}$$$.

Integrera termvis:

$${\color{red}{\int{\left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)}\right)d x}}} = {\color{red}{\left(\int{4 \ln{\left(x \right)} d x} - \int{3 \ln{\left(y \right)} d x}\right)}}$$

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=3 \ln{\left(y \right)}$$$:

$$\int{4 \ln{\left(x \right)} d x} - {\color{red}{\int{3 \ln{\left(y \right)} d x}}} = \int{4 \ln{\left(x \right)} d x} - {\color{red}{\left(3 x \ln{\left(y \right)}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:

$$- 3 x \ln{\left(y \right)} + {\color{red}{\int{4 \ln{\left(x \right)} d x}}} = - 3 x \ln{\left(y \right)} + {\color{red}{\left(4 \int{\ln{\left(x \right)} d x}\right)}}$$

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 »).

Alltså,

$$- 3 x \ln{\left(y \right)} + 4 {\color{red}{\int{\ln{\left(x \right)} d x}}}=- 3 x \ln{\left(y \right)} + 4 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=- 3 x \ln{\left(y \right)} + 4 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

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

$$4 x \ln{\left(x \right)} - 3 x \ln{\left(y \right)} - 4 {\color{red}{\int{1 d x}}} = 4 x \ln{\left(x \right)} - 3 x \ln{\left(y \right)} - 4 {\color{red}{x}}$$

Alltså,

$$\int{\left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)}\right)d x} = 4 x \ln{\left(x \right)} - 3 x \ln{\left(y \right)} - 4 x$$

Förenkla:

$$\int{\left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)}\right)d x} = x \left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)} - 4\right)$$

Lägg till integrationskonstanten:

$$\int{\left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)}\right)d x} = x \left(4 \ln{\left(x \right)} - 3 \ln{\left(y \right)} - 4\right)+C$$

$$$\int \left(\ln\left(x^{4}\right) - 3 \ln\left(y\right)\right)\, dx = x \left(4 \ln\left(x\right) - 3 \ln\left(y\right) - 4\right) + C$$$A