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

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

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

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

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

$$2 {\color{red}{\int{\frac{x^{4}}{x - 1} d x}}} = 2 {\color{red}{\int{\left(x^{3} + x^{2} + x + 1 + \frac{1}{x - 1}\right)d x}}}$$

Gliedweise integrieren:

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

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

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

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

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

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

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

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=3$$$ an:

$$\frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \int{\frac{1}{x - 1} d x} + 2 {\color{red}{\int{x^{3} d x}}}=\frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \int{\frac{1}{x - 1} d x} + 2 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=\frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \int{\frac{1}{x - 1} d x} + 2 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

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,

$$\frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 {\color{red}{\int{\frac{1}{x - 1} d x}}} = \frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$\frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 {\color{red}{\int{\frac{1}{u} d u}}} = \frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$\frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}$$

Daher,

$$\int{\frac{2 x^{4}}{x - 1} d x} = \frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \ln{\left(\left|{x - 1}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{2 x^{4}}{x - 1} d x} = \frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \ln{\left(\left|{x - 1}\right| \right)}+C$$

$$$\int \frac{2 x^{4}}{x - 1}\, dx = \left(\frac{x^{4}}{2} + \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \ln\left(\left|{x - 1}\right|\right)\right) + C$$$A