$$$\frac{x \left(x - 9\right) \left(x - 6\right)}{9}$$$ 的积分

该计算器将求出$$$\frac{x \left(x - 9\right) \left(x - 6\right)}{9}$$$的积分/原函数,并显示步骤。

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您的输入

$$$\int \frac{x \left(x - 9\right) \left(x - 6\right)}{9}\, dx$$$

解答

$$$c=\frac{1}{9}$$$$$$f{\left(x \right)} = x \left(x - 9\right) \left(x - 6\right)$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$${\color{red}{\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x}}} = {\color{red}{\left(\frac{\int{x \left(x - 9\right) \left(x - 6\right) d x}}{9}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{x \left(x - 9\right) \left(x - 6\right) d x}}}}{9} = \frac{{\color{red}{\int{\left(x^{3} - 15 x^{2} + 54 x\right)d x}}}}{9}$$

逐项积分:

$$\frac{{\color{red}{\int{\left(x^{3} - 15 x^{2} + 54 x\right)d x}}}}{9} = \frac{{\color{red}{\left(\int{54 x d x} - \int{15 x^{2} d x} + \int{x^{3} d x}\right)}}}{9}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=3$$$

$$\frac{\int{54 x d x}}{9} - \frac{\int{15 x^{2} d x}}{9} + \frac{{\color{red}{\int{x^{3} d x}}}}{9}=\frac{\int{54 x d x}}{9} - \frac{\int{15 x^{2} d x}}{9} + \frac{{\color{red}{\frac{x^{1 + 3}}{1 + 3}}}}{9}=\frac{\int{54 x d x}}{9} - \frac{\int{15 x^{2} d x}}{9} + \frac{{\color{red}{\left(\frac{x^{4}}{4}\right)}}}{9}$$

$$$c=15$$$$$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{{\color{red}{\int{15 x^{2} d x}}}}{9} = \frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{{\color{red}{\left(15 \int{x^{2} d x}\right)}}}{9}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=2$$$

$$\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{5 {\color{red}{\int{x^{2} d x}}}}{3}=\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{5 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}}{3}=\frac{x^{4}}{36} + \frac{\int{54 x d x}}{9} - \frac{5 {\color{red}{\left(\frac{x^{3}}{3}\right)}}}{3}$$

$$$c=54$$$$$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$

$$\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + \frac{{\color{red}{\int{54 x d x}}}}{9} = \frac{x^{4}}{36} - \frac{5 x^{3}}{9} + \frac{{\color{red}{\left(54 \int{x d x}\right)}}}{9}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=1$$$

$$\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 6 {\color{red}{\int{x d x}}}=\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 6 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 6 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

因此,

$$\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x} = \frac{x^{4}}{36} - \frac{5 x^{3}}{9} + 3 x^{2}$$

化简:

$$\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x} = \frac{x^{2} \left(x^{2} - 20 x + 108\right)}{36}$$

加上积分常数:

$$\int{\frac{x \left(x - 9\right) \left(x - 6\right)}{9} d x} = \frac{x^{2} \left(x^{2} - 20 x + 108\right)}{36}+C$$

答案

$$$\int \frac{x \left(x - 9\right) \left(x - 6\right)}{9}\, dx = \frac{x^{2} \left(x^{2} - 20 x + 108\right)}{36} + C$$$A


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