Integral von $$$- x^{2} \ln\left(x\right)$$$

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=-1$$$ und $$$f{\left(x \right)} = x^{2} \ln{\left(x \right)}$$$ an:

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

Für das Integral $$$\int{x^{2} \ln{\left(x \right)} d x}$$$ verwenden Sie die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Seien $$$\operatorname{u}=\ln{\left(x \right)}$$$ und $$$\operatorname{dv}=x^{2} dx$$$.

Dann gilt $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{x^{2} d x}=\frac{x^{3}}{3}$$$ (Rechenschritte siehe »).

Das Integral wird zu

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(x \right)} = x^{2}$$$ an:

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

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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