Integral von $$$y \left(x^{2} \ln\left(x y\right) + 1\right)$$$ nach $$$x$$$

Bestimme $$$\int y \left(x^{2} \ln\left(x y\right) + 1\right)\, dx$$$.

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

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:

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

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

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

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

Somit,

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

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:

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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