Integral von $$$\frac{x - 1}{x^{2} + 2 x + 3}$$$

Bestimme $$$\int \frac{x - 1}{x^{2} + 2 x + 3}\, dx$$$.

Schreibe den linearen Term als $$$x - 1=x\color{red}{+1-1}-1=x+1-2$$$ um und zerlege den Ausdruck.:

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

Gliedweise integrieren:

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

Sei $$$u=x^{2} + 2 x + 3$$$.

Dann $$$du=\left(x^{2} + 2 x + 3\right)^{\prime }dx = \left(2 x + 2\right) dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\left(2 x + 2\right) dx = du$$$.

Das Integral lässt sich umschreiben als

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

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:

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

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

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

Zur Erinnerung: $$$u=x^{2} + 2 x + 3$$$:

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

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

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

Quadrat ergänzen (Schritte siehe »): $$$x^{2} + 2 x + 3 = \left(x + 1\right)^{2} + 2$$$:

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

Sei $$$u=x + 1$$$.

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

Das Integral lässt sich umschreiben als

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

Sei $$$v=\frac{\sqrt{2} u}{2}$$$.

Dann $$$dv=\left(\frac{\sqrt{2} u}{2}\right)^{\prime }du = \frac{\sqrt{2}}{2} du$$$ (die Schritte sind » zu sehen), und es gilt $$$du = \sqrt{2} dv$$$.

Das Integral wird zu

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

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

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

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

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

Zur Erinnerung: $$$v=\frac{\sqrt{2} u}{2}$$$:

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

Zur Erinnerung: $$$u=x + 1$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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