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

Bestäm $$$\int w^{2} \ln\left(w\right)\, dw$$$.

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

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

Då gäller $$$\operatorname{du}=\left(\ln{\left(w \right)}\right)^{\prime }dw=\frac{dw}{w}$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{w^{2} d w}=\frac{w^{3}}{3}$$$ (stegen kan ses »).

Integralen blir

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

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

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

Tillämpa potensregeln $$$\int w^{n}\, dw = \frac{w^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=2$$$:

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

Alltså,

$$\int{w^{2} \ln{\left(w \right)} d w} = \frac{w^{3} \ln{\left(w \right)}}{3} - \frac{w^{3}}{9}$$

Förenkla:

$$\int{w^{2} \ln{\left(w \right)} d w} = \frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}$$

Lägg till integrationskonstanten:

$$\int{w^{2} \ln{\left(w \right)} d w} = \frac{w^{3} \left(3 \ln{\left(w \right)} - 1\right)}{9}+C$$

$$$\int w^{2} \ln\left(w\right)\, dw = \frac{w^{3} \left(3 \ln\left(w\right) - 1\right)}{9} + C$$$A