Integral von $$$x^{2} \operatorname{atan}{\left(x \right)}$$$

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

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

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

Dann gilt $$$\operatorname{du}=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}$$$ (Rechenschritte siehe ») und $$$\operatorname{v}=\int{x^{2} d x}=\frac{x^{3}}{3}$$$ (Rechenschritte siehe »).

Somit,

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

Den Integranden vereinfachen:

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

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

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

Da der Grad des Zählers mindestens so groß ist wie der des Nenners, führen Sie eine Polynomdivision durch (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

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

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

Sei $$$u=x^{2} + 1$$$.

Dann $$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (die Schritte sind » zu sehen), und es gilt $$$x dx = \frac{du}{2}$$$.

Das Integral wird zu

$$\frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{{\color{red}{\int{\frac{x}{x^{2} + 1} d x}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{{\color{red}{\int{\frac{1}{2 u} d u}}}}{3}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(u \right)} = \frac{1}{u}$$$ an:

$$\frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{{\color{red}{\int{\frac{1}{2 u} d u}}}}{3} = \frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{3}$$

Das Integral von $$$\frac{1}{u}$$$ ist $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6} = \frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Zur Erinnerung: $$$u=x^{2} + 1$$$:

$$\frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} = \frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{\ln{\left(\left|{{\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}}{6}$$

Daher,

$$\int{x^{2} \operatorname{atan}{\left(x \right)} d x} = \frac{x^{3} \operatorname{atan}{\left(x \right)}}{3} - \frac{x^{2}}{6} + \frac{\ln{\left(x^{2} + 1 \right)}}{6}$$

Vereinfachen:

$$\int{x^{2} \operatorname{atan}{\left(x \right)} d x} = \frac{2 x^{3} \operatorname{atan}{\left(x \right)} - x^{2} + \ln{\left(x^{2} + 1 \right)}}{6}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{x^{2} \operatorname{atan}{\left(x \right)} d x} = \frac{2 x^{3} \operatorname{atan}{\left(x \right)} - x^{2} + \ln{\left(x^{2} + 1 \right)}}{6}+C$$

$$$\int x^{2} \operatorname{atan}{\left(x \right)}\, dx = \frac{2 x^{3} \operatorname{atan}{\left(x \right)} - x^{2} + \ln\left(x^{2} + 1\right)}{6} + C$$$A