Integral von $$$\frac{2 x^{4}}{x^{4} - 1}$$$

Bestimme $$$\int \frac{2 x^{4}}{x^{4} - 1}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = \frac{x^{4}}{x^{4} - 1}$$$ an:

$${\color{red}{\int{\frac{2 x^{4}}{x^{4} - 1} d x}}} = {\color{red}{\left(2 \int{\frac{x^{4}}{x^{4} - 1} d x}\right)}}$$

Forme den Bruch um und zerlege ihn:

$$2 {\color{red}{\int{\frac{x^{4}}{x^{4} - 1} d x}}} = 2 {\color{red}{\int{\left(1 + \frac{1}{x^{4} - 1}\right)d x}}}$$

Gliedweise integrieren:

$$2 {\color{red}{\int{\left(1 + \frac{1}{x^{4} - 1}\right)d x}}} = 2 {\color{red}{\left(\int{1 d x} + \int{\frac{1}{x^{4} - 1} d x}\right)}}$$

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

$$2 \int{\frac{1}{x^{4} - 1} d x} + 2 {\color{red}{\int{1 d x}}} = 2 \int{\frac{1}{x^{4} - 1} d x} + 2 {\color{red}{x}}$$

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

$$2 x + 2 {\color{red}{\int{\frac{1}{x^{4} - 1} d x}}} = 2 x + 2 {\color{red}{\int{\left(- \frac{1}{2 \left(x^{2} + 1\right)} - \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x - 1\right)}\right)d x}}}$$

Gliedweise integrieren:

$$2 x + 2 {\color{red}{\int{\left(- \frac{1}{2 \left(x^{2} + 1\right)} - \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x - 1\right)}\right)d x}}} = 2 x + 2 {\color{red}{\left(\int{\frac{1}{4 \left(x - 1\right)} d x} - \int{\frac{1}{4 \left(x + 1\right)} d x} - \int{\frac{1}{2 \left(x^{2} + 1\right)} d x}\right)}}$$

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

$$2 x + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - 2 \int{\frac{1}{4 \left(x + 1\right)} d x} - 2 {\color{red}{\int{\frac{1}{2 \left(x^{2} + 1\right)} d x}}} = 2 x + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - 2 \int{\frac{1}{4 \left(x + 1\right)} d x} - 2 {\color{red}{\left(\frac{\int{\frac{1}{x^{2} + 1} d x}}{2}\right)}}$$

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

$$2 x + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - 2 \int{\frac{1}{4 \left(x + 1\right)} d x} - {\color{red}{\int{\frac{1}{x^{2} + 1} d x}}} = 2 x + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - 2 \int{\frac{1}{4 \left(x + 1\right)} d x} - {\color{red}{\operatorname{atan}{\left(x \right)}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(x \right)} = \frac{1}{x + 1}$$$ an:

$$2 x - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - 2 {\color{red}{\int{\frac{1}{4 \left(x + 1\right)} d x}}} = 2 x - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - 2 {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{4}\right)}}$$

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

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

Somit,

$$2 x - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x + 1} d x}}}}{2} = 2 x - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$2 x - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = 2 x - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

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

$$2 x - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x} = 2 x - \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2} - \operatorname{atan}{\left(x \right)} + 2 \int{\frac{1}{4 \left(x - 1\right)} d x}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(x \right)} = \frac{1}{x - 1}$$$ an:

$$2 x - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \operatorname{atan}{\left(x \right)} + 2 {\color{red}{\int{\frac{1}{4 \left(x - 1\right)} d x}}} = 2 x - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \operatorname{atan}{\left(x \right)} + 2 {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{4}\right)}}$$

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 wird zu

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

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

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

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

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

Daher,

$$\int{\frac{2 x^{4}}{x^{4} - 1} d x} = 2 x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \operatorname{atan}{\left(x \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{2 x^{4}}{x^{4} - 1} d x} = 2 x + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \operatorname{atan}{\left(x \right)}+C$$

$$$\int \frac{2 x^{4}}{x^{4} - 1}\, dx = \left(2 x + \frac{\ln\left(\left|{x - 1}\right|\right)}{2} - \frac{\ln\left(\left|{x + 1}\right|\right)}{2} - \operatorname{atan}{\left(x \right)}\right) + C$$$A