$$$2 x^{3} \left(3 x - 2\right)$$$ 的积分
您的输入
求$$$\int 2 x^{3} \left(3 x - 2\right)\, dx$$$。
解答
输入已重写为:$$$\int{2 x^{3} \left(3 x - 2\right) d x}=\int{x^{3} \left(6 x - 4\right) d x}$$$。
化简被积函数:
$${\color{red}{\int{x^{3} \left(6 x - 4\right) d x}}} = {\color{red}{\int{2 x^{3} \left(3 x - 2\right) d x}}}$$
对 $$$c=2$$$ 和 $$$f{\left(x \right)} = x^{3} \left(3 x - 2\right)$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$${\color{red}{\int{2 x^{3} \left(3 x - 2\right) d x}}} = {\color{red}{\left(2 \int{x^{3} \left(3 x - 2\right) d x}\right)}}$$
Expand the expression:
$$2 {\color{red}{\int{x^{3} \left(3 x - 2\right) d x}}} = 2 {\color{red}{\int{\left(3 x^{4} - 2 x^{3}\right)d x}}}$$
逐项积分:
$$2 {\color{red}{\int{\left(3 x^{4} - 2 x^{3}\right)d x}}} = 2 {\color{red}{\left(- \int{2 x^{3} d x} + \int{3 x^{4} d x}\right)}}$$
对 $$$c=2$$$ 和 $$$f{\left(x \right)} = x^{3}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$2 \int{3 x^{4} d x} - 2 {\color{red}{\int{2 x^{3} d x}}} = 2 \int{3 x^{4} d x} - 2 {\color{red}{\left(2 \int{x^{3} d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=3$$$:
$$2 \int{3 x^{4} d x} - 4 {\color{red}{\int{x^{3} d x}}}=2 \int{3 x^{4} d x} - 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=2 \int{3 x^{4} d x} - 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$
对 $$$c=3$$$ 和 $$$f{\left(x \right)} = x^{4}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$:
$$- x^{4} + 2 {\color{red}{\int{3 x^{4} d x}}} = - x^{4} + 2 {\color{red}{\left(3 \int{x^{4} d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$,其中 $$$n=4$$$:
$$- x^{4} + 6 {\color{red}{\int{x^{4} d x}}}=- x^{4} + 6 {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=- x^{4} + 6 {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$
因此,
$$\int{x^{3} \left(6 x - 4\right) d x} = \frac{6 x^{5}}{5} - x^{4}$$
化简:
$$\int{x^{3} \left(6 x - 4\right) d x} = \frac{x^{4} \left(6 x - 5\right)}{5}$$
加上积分常数:
$$\int{x^{3} \left(6 x - 4\right) d x} = \frac{x^{4} \left(6 x - 5\right)}{5}+C$$
答案
$$$\int 2 x^{3} \left(3 x - 2\right)\, dx = \frac{x^{4} \left(6 x - 5\right)}{5} + C$$$A