Integral von $$$\frac{x \left(x - 9\right) \left(x - 6\right)}{9}$$$
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Ihre Eingabe
Bestimme $$$\int \frac{x \left(x - 9\right) \left(x - 6\right)}{9}\, dx$$$.
Lösung
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{9}$$$ und $$$f{\left(x \right)} = x \left(x - 9\right) \left(x - 6\right)$$$ an:
$${\color{red}{\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x}}} = {\color{red}{\left(\frac{\int{x \left(x - 9\right) \left(x - 6\right) d x}}{9}\right)}}$$
Expand the expression:
$$\frac{{\color{red}{\int{x \left(x - 9\right) \left(x - 6\right) d x}}}}{9} = \frac{{\color{red}{\int{\left(x^{3} - 15 x^{2} + 54 x\right)d x}}}}{9}$$
Gliedweise integrieren:
$$\frac{{\color{red}{\int{\left(x^{3} - 15 x^{2} + 54 x\right)d x}}}}{9} = \frac{{\color{red}{\left(\int{54 x d x} - \int{15 x^{2} d x} + \int{x^{3} d x}\right)}}}{9}$$
Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=3$$$ an:
$$\frac{\int{54 x d x}}{9} - \frac{\int{15 x^{2} d x}}{9} + \frac{{\color{red}{\int{x^{3} d x}}}}{9}=\frac{\int{54 x d x}}{9} - \frac{\int{15 x^{2} d x}}{9} + \frac{{\color{red}{\frac{x^{1 + 3}}{1 + 3}}}}{9}=\frac{\int{54 x d x}}{9} - \frac{\int{15 x^{2} d x}}{9} + \frac{{\color{red}{\left(\frac{x^{4}}{4}\right)}}}{9}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=15$$$ und $$$f{\left(x \right)} = x^{2}$$$ an:
$$\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{{\color{red}{\int{15 x^{2} d x}}}}{9} = \frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{{\color{red}{\left(15 \int{x^{2} d x}\right)}}}{9}$$
Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:
$$\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{5 {\color{red}{\int{x^{2} d x}}}}{3}=\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{5 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}}{3}=\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{5 {\color{red}{\left(\frac{x^{3}}{3}\right)}}}{3}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=54$$$ und $$$f{\left(x \right)} = x$$$ an:
$$\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + \frac{{\color{red}{\int{54 x d x}}}}{9} = \frac{x^{4}}{36} - \frac{5 x^{3}}{9} + \frac{{\color{red}{\left(54 \int{x d x}\right)}}}{9}$$
Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=1$$$ an:
$$\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 6 {\color{red}{\int{x d x}}}=\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 6 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 6 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Daher,
$$\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x} = \frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 3 x^{2}$$
Vereinfachen:
$$\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x} = \frac{x^{2} \left(x^{2} - 20 x + 108\right)}{36}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x} = \frac{x^{2} \left(x^{2} - 20 x + 108\right)}{36}+C$$
Antwort
$$$\int \frac{x \left(x - 9\right) \left(x - 6\right)}{9}\, dx = \frac{x^{2} \left(x^{2} - 20 x + 108\right)}{36} + C$$$A