Integralen av $$$\ln^{2}\left(3 x\right)$$$

Bestäm $$$\int \ln^{2}\left(3 x\right)\, dx$$$.

Låt $$$u=3 x$$$ vara.

Då $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{3}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{3}$$$ och $$$f{\left(u \right)} = \ln{\left(u \right)}^{2}$$$:

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

För integralen $$$\int{\ln{\left(u \right)}^{2} d u}$$$, använd partiell integration $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Låt $$$\operatorname{g}=\ln{\left(u \right)}^{2}$$$ och $$$\operatorname{dv}=du$$$.

Då gäller $$$\operatorname{dg}=\left(\ln{\left(u \right)}^{2}\right)^{\prime }du=\frac{2 \ln{\left(u \right)}}{u} du$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d u}=u$$$ (stegen kan ses »).

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=2$$$ och $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

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

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 »).

Integralen kan omskrivas som

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

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

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

Kom ihåg att $$$u=3 x$$$:

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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