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

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

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 »).

Das Integral wird zu

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

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

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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