Integral von $$$\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1$$$

Bestimme $$$\int \left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)\, dx$$$.

Gliedweise integrieren:

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

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

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

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

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=2$$$ an:

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

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

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

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

Um das Integral $$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}$$$ zu berechnen, wende die partielle Integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$ auf das Integral $$$\int{\frac{1}{x^{2} + 1} d x}$$$ an.

Seien $$$\operatorname{u}=\frac{1}{x^{2} + 1}$$$ und $$$\operatorname{dv}=dx$$$.

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

Somit,

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}=\frac{1}{x^{2} + 1} \cdot x-\int{x \cdot \left(- \frac{2 x}{\left(x^{2} + 1\right)^{2}}\right) d x}=\frac{x}{x^{2} + 1} - \int{\left(- \frac{2 x^{2}}{\left(x^{2} + 1\right)^{2}}\right)d x}$$

Konstante ausklammern:

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

Schreibe den Zähler des Integranden als $$$x^{2}=x^{2}{\color{red}{+1}}{\color{red}{-1}}$$$ um und zerlege:

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

Teile die Integrale auf:

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

Damit erhalten wir die folgende einfache lineare Gleichung bezüglich des Integrals:

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

Durch Lösen erhalten wir, dass

$$\int{\frac{1}{\left(x^{2} + 1\right)^{2}} d x}=\frac{x}{2 \left(x^{2} + 1\right)} + \frac{\int{\frac{1}{x^{2} + 1} d x}}{2}$$

Daher,

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

Das Integral von $$$\frac{1}{x^{2} + 1}$$$ ist $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

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

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

Das Integral von $$$\frac{1}{x^{2} + 1}$$$ ist $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

Daher,

$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{x^{3}}{3} - x - \frac{x}{2 \left(x^{2} + 1\right)} + \frac{5 \operatorname{atan}{\left(x \right)}}{2}$$

Vereinfachen:

$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)d x} = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)}+C$$

$$$\int \left(\frac{x^{6}}{\left(x^{2} + 1\right)^{2}} + 1\right)\, dx = \frac{- 3 x + \left(x^{2} + 1\right) \left(2 x^{3} - 6 x + 15 \operatorname{atan}{\left(x \right)}\right)}{6 \left(x^{2} + 1\right)} + C$$$A